Construct of the form
n=0 anxn
where x is a formal variable and the an are coefficients (usually coming from a field, but sometimes from more exotic objects. Using the standard rules of algebra we can perform formal operations on the series, like addition and multiplication; others may be possible, depending on where we get our coefficients from.

Sometimes we'll use a power series about x0, by replacing "x" above with "(x-x0)" throughout.

In analysis we usually also demand that our power series converge in some neighbourhood.

All Taylor series (and MacLaurin series) are power series.

Power series are most important in the context of complex analysis, and most of the time the term power series is used to refer to a complex power series (real power series are best considered as a special case of complex power series).

The first important property of power series is that they converge on some disc, so the define a function on a nice and sensible domain. The idea of the proof is quite simple, but it uses extra (simple) machinery of uniform convergence to show that the function is not just well-defined on the disc but also continuous.
The second important property is that a function defined by a power series is in fact not only continuous but also complex differentiable. Thus power series give analytic functions, which are extremely nice objects. There is also a converse to this: Taylor's theorem for complex functions states that any analytic function can be expressed locally as a power series (note that this does not quite hold for real functions, even if they are infinitely differentiable)

Proposition:
Given a sequence of complex numbers an, n = 1, 2, 3, ..., there is a non-negative real number R (which may take the values 0 and ∞), called the radius of convergence, such that the series Σ anzn (this and all subsequent sums are taken from n = 0 to ∞) converges for all z in the disc A = {z ∈ C : |z| < R}. The series converges locally uniformly on A, and the function defined by the series is continuous on A.

Proof:
Suppose that the Σ anwn converges for some w ∈ C. Then the sequence anwn must be bounded by some constant M. Let k be any real number < 1, and B the disc {z ∈ C : |z| < k|w|}. Then

Σ supz∈E|anzn| < Σ |anknwn| < Σ Bkn = M/(1-k)

so by the Weierstrass M-test Σ anzn converges uniformly on B.
Thus if we let R = sup {|z| : Σ anzn converges} we have that Σ anzn converges uniformly on the disc |z| < r for any r < R. Hence for any z ∈ A we have that Σ anzn converges uniformly on an open set containing z, which is what we mean by saying that Σ anzn converges locally uniformly on A.
Since Σ anzn is a locally uniform limit of continuous functions it is itself continuous.

Proposition:
Let f(z) = Σ anzn on the disc A. Then f is complex differentiable on A, and the derivative is the term-wise derivative Df(z) = Σ nanzn-1.

Proof:
First of all we note that Df has the same radius of convergence as f, and so is defined and continuous on A. We can do this either by more or less repeating the proof above, or we can use Hadamard's formula for the radius of convergence. We also note that applying this to Df gives that D2f(z) too is well-defined and continuous on A.
Now for z ∈ A and h sufficiently small

|f(z+h) - f(z) - hDf(z)| = |Σ an(z+h)n - anzn - hnanzn-1| ≤ Σ |an(z+h)n - anzn - hnanzn-1| < Σ h2n(n-1)(z+h)n-2 = h2D2f(z+h)

where we have used the binomial theorem together with the estimate n+2Ck+2 < n(n-1)nCk in the last inequality. Since D2f is continuous at z we therefore get

|(f(z+h) - f(z))/h - Df(z)| → 0 as h → 0

which implies that f'(z) exists and equals Df(z).

Corollary:
A power series f(z) is infinitely differentiable inside its radius of convergence, and f(n)(z) = Dnf(z).

Proof:
Since Dnf(z) is itself a power series we can use induction.

Log in or register to write something here or to contact authors.