Power Series, Extending an Analytic Function

Extending an Analytic Function

If f is analytic, and we know the value of f at c, and at points arbitrarily close to c, then we know f over a neighborhood of c. And once f is established on a neighborhood of c, it can be extended to its entire domain. If you know how f behaves as it approaches c, from any direction, then you know f everywhere. The extension of an analytic function is unique.

Let a sequence of points approach the limit point c in the complex plane. If infinitely many points in the sequence are zeros, then f(c) = 0 by continuity. The zeros of an analytic function are isolated, so we have a contradiction. The only way out is to set f equal to the 0 function.

Let two analytic functions f and g attain the same values on c, and points arbitrarily close to c. Each has a power series, with a circle of convergence centered at c. Let r be the lesser of the two radii. Thus the power series for f-g converges out to a radius of r. What does this power series look like? Since f and g are equal on the sequence approaching c, f-g is 0 as it approaches c. By the above, f-g = 0, and the series for f and the series for g are the same. This means f = g, at least on the circle centered at c, with radius r.

There is a good chance that convergence stopped at r because of a pole. For example, 1/(1-z) converges up to the unit circle, and no farther, but the function is actually defined on the entire plane, except for the pole at 1. So - f and g are equal on a disk of radius r, but are they equal everywhere, or do they diverge later on?

One could set f = 1 above the real line and f = -1 below the real line. This is analytic wherever it is defined, yet the behavior of f below the real line does not affect or predict the behavior of f above the real line. Disconnected domains can present independent functions. With this in mind, we will try to predict the behavior of f across its connected domain. That's the best we can do.

If you are a topologist, you might wonder about connected verses path connected in the following proof. Well - the domain of an analytic function is always an open set in the plane, hence connected and path conected are the same.

Let d be a point in the same connected component as c, and suppose f(d) ≠ g(d). Draw a path from c to d. Let r(t) be the radius of convergence of the power series of f, or g, whichever is smaller, at a point t along the path. For each t, establish an open disk of radius ½r(t), centered at t. These open sets cover the path. Since the path is compact, a finite subcover will do. This establishes a chain of intersecting open disks, each half as large as it needs to be to assure convergence of f and g.

Draw a finite chain of segments conecting each disk's center to the next. This begins with c and ends with d. How much wiggle room do we have on either side of this chain for convergence? Let s be the smallest radius of the open disks in the chain. If adjacent disks, of size s, barely overlap, convergence extends from their centers out to a distance of 2s, Which is sqrt(3)s from the midpoint of the connecting segment. Along the segment, we can move a distance s on either side, and f and g are analytic. The chain has thickened to a road of width 2s, with a few sharp turns, and a yello line down the middle.

Place a series of circles along this chain, each having radius s, so that their centers are s units apart or less. In other words, the center of the "next" circle sits on the circumference of the previous circle.

Start at c, where f = g out to a distance of 2s. Move to the center of the next circle, which is s units away from c. Since f = g near this point, f = g out to a distance of 2s. Move to the center of the third circle, a distance of s from the second center. Once again f = g, extending out to a distance of 2s. This embraces the center of the fourth circle, and so on, all the way out to d. Once we reach d, f(d) = g(d).

If d is in the same connected domain as c, i.e. there is a path from c to d, the behavior of f at c completely determines the behavior of f at d.

If f is 0 near c, 0 spreads down the chain, and f is 0 everywhere.

In summary, an analytic function is completely defined, across its connected domain, by a sequence approaching a limit point c, or by its derivatives at c.

Real Analytic

If a real function f(x) can be extended to an analytic function f(z), f is called real analytic. A function is real analytic iff it agrees with its taylor series for some interval about the point of expansion. A sufficient condition for real analytic in a neighborhood of radius r is the presence of a constant m, such that the k^th derivative of f(x) ≤ m^k for all x in this neighborhood. The error term of the k^th taylor approximation is the k^th derivative of f, evaluated at some c in our neighborhood, times x^k/k!. This is bounded by (mx)^k/k!, which approaches 0 for large k, hence the taylor series agrees with f.

The only possible analytic extension of E^x in the complex plane is E^z. It is defined by the taylor series of E^x; just replace x with z.

If a relationship among analytic functions holds for real x, it extends to the entire domain. The function sin(z)² + cos(z)² -1 cannot be 0 on the x axis and nonzero elsewhere, so the familiar trig identity must be true over the complex plane.

When f(x) is Real

If the domain of an analytic function f includes a piece of the x axis, f is real (on the x axis) iff f conjugate of z = f of z conjugate. Conjugate both sides of this equation to get the equivalent condition f conjugate of z conjugate = f(z).

If the condition holds then apply it to a real number x, and f conjugate of x = f(x), hence f(x) is real.

To show the converse, verify that f conjugate of z conjugate is a differentiable function iff f is differentiable at z. Write the difference quotient and take the limit as h approaches 0, and find the conjugate of f′, evaluated at the conjugate of z. I'll leave the algebra to you.

Let g = f on the upper half plane, and f conjugate of z conjugate on the lower half plane. Note that this definition is consistent on the x axis. Thanks to the above paragram, g is analytic above and below the x axis.

Let p be a real number, with g(p) = r. By assumption, r is also real. As we approach p from above, f approaches r, which means f conjugate also approaches r. Thus g approaches r from above and below, and g is continuous at p.

Since f is analytic, it is differentiable at p. As you recall, f′(p) times 1 gives the change in f as we move away from p in the x direction. Since f is real on the x axis, the change in f is real, and the derivative is real. Therefore g is differentiable from above, and g′(p) is a real number. The difference quotient from below approaches the same real value, so g is differentiable on the real line.

The extension of an analytic function is unique, so start with f on the upper half plane and extend to the lower half. This implies f = g = f conjugate of z conjugate.

For another proof, note that f′(x) is real for x real, and so on for the remaining derivatives. The power series consists of real coefficients. We may conjugate z, or the partial sum of the power series out to the n^th term; the result is the same. Since partial sums are equal under pre and post conjugation, the same is true of f. Thus f conjugate of z = f of z conjugate.

To illustrate, let's apply this to E^z. The images E^a+bi and E^a-bi both lie on a circle, centered at the origin, with radius E^a. To find the first image, move around the circle in a counterclockwise direction, through an angle of b. To find the second image, move in a clockwise direction. The two images are symmetric about the real axis. In other words, they are conjugate.