Analytic Functions

What It Means to Be Analytic

The previous tutorial showed that having derivatives of every order does not, by itself, guarantee that a function equals its Taylor series. Analyticity adds exactly the local representation that was missing: near each point, the function must agree with a convergent power series. This is a local condition. The series used near one point need not be the same series, or have the same coefficients, as the one used near another.

The radius \(r\) in this definition only needs to be positive; it need not be the full radius of convergence of the displayed series. Also, convergence alone is not enough: the sum must agree with \(f\) throughout a neighborhood of the center.

The definition implies that an analytic function has derivatives of every order at its center, with coefficients determined by those derivatives. This is a consequence of the term-by-term differentiation theorem for power series, not an additional assumption in the definition.

Proof. By analyticity, the displayed power series represents \(f\) for \(|x-a|<r\), for some \(r>0\). The term-by-term differentiation theorem for power series says that the differentiated series represents the derivative inside the radius of convergence and has the same radius of convergence as the original series. Applying that theorem repeatedly shows that \(f\) has derivatives of every order there. Each derivative is itself represented by a power series, and the continuity theorem for power-series sums shows that each derivative is continuous inside the radius.

After differentiating \(k\) times, evaluate the resulting series at \(x=a\). Every term vanishes except the constant term arising from \(c_k(x-a)^k\). That term is \(k!c_k\), so \(f^{(k)}(a)=k!c_k\). Thus \(c_k=f^{(k)}(a)/k!\) for each \(k\), which identifies the original series as the Taylor series of \(f\) at \(a\). \(\square\)

This result gives a useful way to recognize analyticity: one can first identify a power series representing the function near a point, then read off all of its derivatives there. Conversely, the Taylor Series tutorial established the coefficients that any such representation must have. The essential extra condition is that the Taylor series actually converges to the function throughout a neighborhood.

Worked Examples

Sums and Products of Analytic Functions

Analyticity is preserved by basic algebraic operations. For products, the relevant fact is that two absolutely convergent power series can be multiplied by forming the Cauchy product. The absolute convergence needed here holds at every point strictly inside the radii of convergence.

Proof. Since \(f\) and \(g\) are analytic at \(a\), there are positive radii \(r_f,r_g\) and coefficients \(u_n,v_n\) such that, on the respective neighborhoods, $$ f(x)=\sum_{n=0}^{\infty}u_n(x-a)^n, \qquad g(x)=\sum_{n=0}^{\infty}v_n(x-a)^n. $$ Let \(r=\min\{r_f,r_g\}\). For \(|x-a|<r\), both series converge absolutely. Their sum is represented there by $$ f(x)+g(x)=\sum_{n=0}^{\infty}(u_n+v_n)(x-a)^n, $$ so \(f+g\) is analytic at \(a\).

For the product, define \(w_n=\sum_{j=0}^{n}u_jv_{n-j}\). For each \(|x-a|<r\), absolute convergence gives $$ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} |u_jv_k|\,|x-a|^{j+k} = \left(\sum_{j=0}^{\infty}|u_j|\,|x-a|^j\right) \left(\sum_{k=0}^{\infty}|v_k|\,|x-a|^k\right) <\infty. $$ Thus the terms can be grouped by the total index \(n=j+k\). This yields $$ f(x)g(x) =\sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}u_jv_{n-j}\right)(x-a)^n =\sum_{n=0}^{\infty}w_n(x-a)^n. $$ The product is therefore represented by a convergent power series in a neighborhood of \(a\), so it is analytic there. If \(f\) and \(g\) are analytic on \(U\), this argument applies at each \(a\in U\), proving the final assertion. \(\square\)

Why the Local View Matters

Analyticity is defined point by point, and the allowable neighborhood can change from point to point. In the square-root example, the binomial expansion centered at \(a>0\) was established for \(|x-a|<a\); that radius shrinks as the center approaches \(0\). Nothing in the definition requires one fixed radius to work across an entire interval.

Another important distinction is between analytic and infinitely differentiable. The Taylor Representation Theorem proves that analytic functions are infinitely differentiable, but the converse is false. The smooth function from the previous tutorial, whose derivatives at \(0\) all vanish although the function is positive away from \(0\), cannot be represented near \(0\) by its Taylor series. Its Taylor series is identically zero, so it does not agree with the function on any neighborhood of \(0\).

The sum-and-product theorem is useful because it lets analyticity be built from known local power-series representations without calculating derivatives of every order. For more complicated operations, one must still check the relevant hypotheses and verify that a resulting series converges and represents the function locally. Analyticity is a statement about representation, not merely about formal algebra with coefficients.