Analytic Continuation

3.1.3 Analytic Continuation

Analytic functions have many important properties that you may find in any text on complex analysis. A good summary for the purpose of numerical mathematics is found in the first chapter of Stenger [332]. Two important properties are contained in the following lemma.

We remark that the region of analyticity of a function f (z) is an open set. If we say that f (z) is analytic on a closed real interval, it means that there exists an open set in C that contains this interval, where f (z) is analytic.

Lemma 3.1.7.

An analytic function can only have a finite number of zeros in a compact subset of the region of analyticity, unless the function is identically zero. Suppose that two functions f 1 and f 2 are analytic in regions D 1 and D 2 , respectively. Suppose that D 1 ∩D 2 contains an interval throughout which f 1 (z) =f 2 (z). Then f 1 (z) =

f 2 (z) in the intersection D 1 ∩D 2 .

172 Chapter 3. Series, Operators, and Continued Fractions Proof. We refer, for the first part, to any text on complex analysis. Here we closely follow

Titchmarsh [351]. The second part follows by the application of the first part to the function

f 1 −f 2 .

A consequence of this is known as the permanence of functional equations. That is, in order to prove the validity of a functional equation (or “a formula for a function”) in a region of the complex plane, it may be sufficient to prove its validity in (say) an interval of the real axis, under the conditions specified in the lemma.

Example 3.1.8 (The Permanence of Functional Equations). We know from elementary real analysis that the functional equation

e (p +q)z

pz =e qz e , (p, q ∈ R),

holds for all z ∈ R. We also know that all three functions involved are analytic for all z ∈ C. Set D 1 =D 2 = C in the lemma, and let “the interval” be any compact interval of R. The

lemma then tells us that the displayed equation holds for all complex z. The right- and left-hand sides then have identical power series. Applying the convo- lution formula and matching the coefficients of z n , we obtain

n ! p j = n = q −j . n

(p n n + q) p j q n −j

i.e., (p + q)

!(n − j)! This is not a very sensational result. It is more interesting to start from the following

functional equation:

1 + z) p +q

= (1 + z) q ( 1 + z) .

The same argumentation holds, except that—by the discussion around Table 3.1.1—D 1 ,D 2 should be equal to the complex plane with a cut from −1 to −∞, and that the Maclaurin

series is convergent in the unit disk only. We obtain the equations

(They can also be proved by induction, but it is not needed.) This sequence of algebraic identities, where each identity contains a finite number of terms, is equivalent to the above functional equation.

We shall see that this observation is useful for motivating certain “symbolic compu- tations ” with power series, which can provide elegant derivations of useful formulas in numerical mathematics.

Now we may consider the aggregate of values of f 1 (z) and f 2 (z) at points interior to D 1 or D 2 as a single analytic function f . Thus f is analytic in the union D 1 ∪D 2 , and

f (z) =f 1 (z) in D 1 , f (z) = f 2 (z) in D 2 .

The function f 2 may be considered as extending the domain in which f 1 is defined, and it is called a (single-valued) analytic continuation of f 1 . In the same way, f 1 is an

3.1. Some Basic Facts about Series 173 analytic continuation of f 2 . Analytic continuation denotes both this process of extending

the definition of a given function and the result of the process. We shall see examples of this, e.g., in Sec. 3.1.4. Under certain conditions the analytic continuation is unique.

Theorem 3.1.8.

Suppose that a region D is overlapped by regions D 1 ,D 2 , and that (D 1 ∩D 2 ) ∩D contains an interval. Let f be analytic in D, let f 1 be an analytic continuation of f to D 1 ,

and let f 2 be an analytic continuation of f to D 2 so that

f (z) =f 1 (z) =f 2 (z) in (D 1 ∩D 2 ) ∩ D. Then either of these functions provides a single-valued analytic continuation of f to D 1 ∩D 2 .

The results of the two processes are the same. Proof. Since f 1 −f 2 is analytic in D 1 ∩D 2 , and f 1 −f 2 = 0 in the set (D 1 ∩D 2 ) ∩ D,

which contains an interval, it follows from Lemma 3.1.7 that f 1 (z) =f 2 (z) in D 1 ∩D 2 , which proves the theorem.

If the set (D 1 ∩D 2 ) ∩ D is void, the conclusion in the theorem may not be valid. We may still consider the aggregate of values as a single analytic function, but this function can

be multivalued in D 1 ∩D 2 .

Example 3.1.9.

For |x| < 1 the important formula

arctan x =

easily follows from the expansions in Table 3.1.1. The function arctan x has an analytic continuation as single-valued functions in the complex plane with cuts along the imaginary axis from i to ∞ and from −i to −∞. It follows from the theorem that “the important formula” is valid in this set.