Power Series, Laurent Series

Laurent Series

In the previous section f was analytic throughout a disk of radius r, centered at the origin. We expressed f as a power series, a0 + a1z + a2z2 + a3z3 + …, and showed this series converges absolutely, for every z inside our circle of radius r. Furthermore, convergence is uniform on any disk whose radius is strictly less than r. You need to be familiar with that proof, or the rest of this page won't make any sense.

Assume f is analytic throughout an annulus, having outer radius r and inner radius s. We don't know anything about f inside the circle of radius s.

Let z be any point inside the annulus. Let y(t) run around the outer circle in a counterclockwise direction, and around the inner circle in a clockwise direction. We already know that the integral of f(y)/(y-z) is equal to f(z)×2πi.

As you recall from the previous section, we set v = z/y. This time, let u = y/z. We will use v when analyzing the contour integral around the outer circle, as we did before, and u for the contour integral around the inner circle. Let's take the outer circle first.

The integrand is f(y)/(y-z). replace 1/(y-z) with 1/y times 1/(1-v), expand 1/(1-v) as a geometric sum with an error term, and integrate around the outer circle, term by term, as we did before. The equation looks something like this.

f(z) = 1/2πi times the contour integral around the outer circle of {
   f(y)/y + f(y)/y2×z + f(y)/y3*z2 + f(y)/y4*z3 + …
   f(y)vn/(y-z) } -
   the integral around the inner circle, which we'll get to later.

Since f is not analytic at 0, we can't replace the coefficients on the powers of z with the corresponding derivatives at 0, as we did before. That just isn't an option. However, we can leave it as a power series in z, and make a few observations.

Review the proof of convergence given in the previous section, and apply it here. Let m be the maximum of |f| over the annulus, and use m and z to bound the norm of the error term beneath a geometric series. Thus the series generated by the outer circle is absolutely convergent at z. Furthermore, it is uniformly convergent on any closed domain contained in the annulus, and stricly inside the outer circle.

Now how about the inner circle? We are subtracting the contour integral of a sum of terms with y-z in the denominator, so instead, change the denominator to z-y, and add the contour integrals. Since u = y/z, replace 1/(z-y) with 1/z times 1/(1-u), expand 1/(1-u) geometrically, and write the following expression for the contour integral around the inner circle.

   1/2πi times the contour integral around the inner circle of {
   f(y)z-1 + f(y)yz-2 + f(y)y2z-3 + f(y)y3z-4 +
   f(y)un/(z-y) }

With y running around the inner circle, and z strictly inside the annulus, and m bounding |f|, the error term goes to 0. This series is absolutely convergent, and uniformly convergent on a domain that is strictly outside the inner circle.

Put this all together and express f(z) as a linear combination of positive and negative powers of z, where the coefficients are contour integrals around the outer and inner circles. When z is raised to a positive power the integral runs around the outer circle, and when z is raised to a negative power the integral runs around the inner circle. The series converges absolutely to f(z), for every z in the annulus. When the domain is bounded away from the inner and outer circles, the series of functions converges uniformly to f.

If there are finitely many terms with negative exponents, i.e. most of the contour integrals around the inner circle are 0, the series is called a laurent series. (biography) The terms are usually presented in order, like this.

a-3z-3 + a-2z-2 + a-1z-1 + a0 + a1z + a2z2 + a3z3 + a4z4 + …

I will sometimes refer to this as a forward laurent series, as opposed to a reverse laurent series, which presents infinitely many negative exponents. Replace z with 1/z in the above example to get a reverse series.

Actually, the adjectives forward and reverse put a block on the series, at the left and right respectively. A finite laurent series is both a forward laurent series and a reverse laurent series. Remember that a power series is a forward laurent series that has no negative exponents.

A bidirectional laurent series runs forever in both directions, and a general laurent series is unconstrained. If we don't know anything about f inside the smaller circle of our annulus, the contour integrals produce a general laurent series.

Special Case

Assume f is analytic throughout the disk of radius r, but we write the general laurent series for f using the annulus between s and r. Since f is analytic throughout, the constant term becomes f(0), and the coefficients on the positive powers of z become the various derivatives of f at 0, divided by k factorial. What happens to the negative powers of z? The coefficient is a contour integral of f(y), times some power of y, running around the inner circle. Yet f(y)yk is analytic throughout the inner circle, so by the Cauchy Goursat theorem, the integral is 0. The negative powers of z drop out, and the power series returns. Thus the power series is a special case of the general laurent series, as computed by the contour integrals, when f happens to be analytic throughout the entire disk.

Linear Combinations

If f and g are analytic throughout our annulus, the general laurent series of f+g is equal to the series for f plus the series for g. Remember that each coefficient is a contour integral, and the integral of (f+g)×w, where w is some other factor, is the integral of fw plus the integral of gw. We can simply add coefficients across the board.

Using similar reasoning, the series for cf, where c is a constant, is c times the series of f. Put this all together, and the laurent operator, from function to series, is indeed a transform from one vector space into another.

Extra Factors of z

If f produces a general laurent series, consider the series for f times z. If k is positive, the coefficient on zk use to be an integral of f(y)/yk+1. Now there is an extra factor of y in the integrand, and the denominator becomes yk. This integral creates the coefficient that use to be on zk-1. thus the coefficients move up, or if you prefer, the exponents on z increase by 1. This is what you'd expect when f is multiplied by z.

But what about the constant term? It is the integral of f(y)/y, but now f brings in another factor of y, so we have the integral of f. Since f is analytic on the annulus, the integral is the same on the inner circle and the outer circle, so apply it to the inner circle, and find the coefficient that use to be on z-1. Once again the exponent has increased, and a-1 has become a0.

finally let k be negative and verify that the coefficient on zk is the coefficient that use to be on zk-1. Multiplying f by z causes the entire series to shift, as you would expect.

I'll leave it to you to reverse this process. Divide f by z, and the new series is a shifted version of the original, such that the coefficient on zk becomes the coefficient on zk-1.

Reciprocal

If f is analytic on an annulus, and generates a certain series, what about g, defined as f(1/z), which is analytic on its own annulus. The circles bounding g have radii that are inverses of the radii of the circles bounding f.

The kth contour integral of f around its outer circle has an integrand of f(y)/yk+1. (I'm not going to worry about the constants here.) Integrate by substitution, replacing y with 1/w. This gives an integrand of -g(w)wk+1/w2. The new contour is the inner circle of g, running in the clockwise (nonstandard) direction. Reverse the direction of integration and drop the minus sign, giving an integrand of g(w)wk-1. Again, this integral runs along the inner circle of g. It is precisely the integral we would normally evaluate to compute the coefficient on z-k. The coefficient of f, on zk, equals the coefficient of g on z-k.

Similar reasoning shows the coefficient of f, on z-k, equals the coefficient of g on zk. If we know the series for f, we can reverse it to find the series that converges to g, and is generated by the contour integrals, using the annulus of g.