Taylor's Theorem
Theorem: if a function f(z) is analytic in an open disk |z - z_0| < R, then for every point z in this disk, f(z) can be represented by the unique, convergent power series:
f(z) = \sum_{n=0}^\infty c_n (z - z_0)^n
where the coefficients are given by the derivatives of the function at the center of the disk:
c_n = \frac{f^{(n)}(z_0)}{n!}
Proof: let f(z) be analytic in the disk D = \{z \in \mathbb{C} \mid |z - z_0| < R\}. We choose an arbitrary point z within this disk. Let |z - z_0| = r. Since z is in an open disk, we can always find a radius \rho such that r < \rho < R. We construct a circular contour C_\rho centered at z_0 with radius \rho. The point z is strictly inside this contour.
Since f(z) is analytic within and on C_\rho, we can apply Cauchy’s Integral Formula to express the value of the function at z:
f(z) = \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(\zeta)}{\zeta-z} \, \mathrm d\zeta
We can expand the kernel of the integral, \frac{1}{\zeta - z}, into a series. We perform an algebraic manipulation by adding and subtracting z_0 in the denominator:
\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \cdot \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}}
The term \frac{z-z_0}{\zeta-z_0} has a magnitude that is strictly less than one. This is because for any \zeta on the contour C_\rho, we have |\zeta - z_0| = \rho, while for the point z, we have |z - z_0| = r. The condition r < \rho ensures that:
\left| \frac{z - z_0}{\zeta - z_0} \right| = \frac{|z - z_0|}{|\zeta - z_0|} = \frac{r}{\rho} < 1
This allows us to expand the term \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}} as a convergent geometric series:
\frac{1}{1 - \frac{z - z_0}{\zeta - z_0}} = \sum_{n=0}^\infty \left( \frac{z - z_0}{\zeta - z_0} \right)^n
Substituting this series back into the expression for the kernel gives:
\frac{1}{\zeta - z} = \frac{1}{\zeta - z_0} \sum_{n=0}^\infty \frac{(z - z_0)^n}{(\zeta - z_0)^n} = \sum_{n=0}^\infty \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}
We must justify substituting this infinite series into Cauchy’s integral formula and interchanging the order of summation and integration. This is permissible if the series converges uniformly with respect to the integration variable \zeta on the contour C_\rho. To show this, we bound the terms of the series.
Since f(\zeta) is analytic on the compact set C_\rho, it is bounded, i.e., there exists a constant M such that |f(\zeta)| \le M for all \zeta \in C_\rho. The absolute value of each term in the series expansion of the integrand is:
\left| f(\zeta) \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}} \right| \le M \frac{|z-z_0|^n}{|\zeta - z_0|^{n+1}} = M \frac{r^n}{\rho^{n+1}}
The series of these upper bounds is a convergent geometric series:
\sum_{n=0}^\infty M \frac{r^n}{\rho^{n+1}} = \frac{M}{\rho} \sum_{n=0}^\infty \left(\frac{r}{\rho}\right)^n
By the Weierstrass M-test, the series for the integrand converges uniformly for \zeta on C_\rho. We can therefore interchange the summation and integration:
\begin{aligned} f(z) &= \frac{1}{2\pi i} \oint_{C_\rho} f(\zeta) \left( \sum_{n=0}^\infty \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}} \right) \mathrm d\zeta \\ &= \sum_{n=0}^\infty \left( (z - z_0)^n \cdot \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \, \mathrm d\zeta \right) \end{aligned}
This expresses f(z) as a power series in (z - z_0). We can identify the coefficients c_n as:
c_n = \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \, \mathrm d\zeta
From the Cauchy Integral Formula for derivatives, we recognize this integral representation as the formula for the n^{th} derivative of f(z) at z_0, divided by n!:
c_n = \frac{f^{(n)}(z_0)}{n!}
This completes the proof of the existence of the series representation.
To demonstrate that this power series representation is unique, let us assume there exists another expansion for f(z) that converges in the same disk:
f(z) = \sum_{n=0}^\infty c_n^\prime (z - z_0)^n
As established by the corollaries of Abel’s theorem (here), the coefficients of any convergent power series representing a function are uniquely determined by the function’s derivatives at the center of expansion. Specifically, for the second series, the coefficients must be given by:
c_n^\prime = \frac{f^{(n)}(z_0)}{n!}
However, we have already established that our original coefficients c_n are also equal to this exact expression. It follows directly that c_n^\prime = c_n for all n \ge 0, confirming that the Taylor expansion of an analytic function is unique.
The expansion of an analytic function in a convergent power series is known as its Taylor series expansion. Taylor’s theorem establishes a one-to-one correspondence between a function analytic in a neighborhood of a point z_0 and a power series centered at that point.
This implies that in complex analysis, the concept of an analytic function (one that is infinitely complex-differentiable) is equivalent to a function that can be locally represented as the sum of a power series. This equivalence does not hold for functions of a real variable.
Furthermore, if a function f(z) is analytic in a domain \mathcal D and z_0 is any point in \mathcal D, the radius of convergence of its Taylor series centered at z_0 is at least as large as the distance from z_0 to the boundary of \mathcal D.
Geometric series analysis
The series \sum_{n=0}^\infty (z - z_0)^n is a example of a power series centered at z_0. As it is also a geometric series, we can determine both its region of convergence and its sum.
To determine the values of z for which the series converges, we can use the D’Alembert’s ratio test. For a series \sum a_n, the test states that the series converges absolutely if the following limit is less than 1:
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1
For our series, the terms are a_n = (z - z_0)^n. We apply the test by forming the ratio of consecutive terms:
\frac{a_{n+1}}{a_n} = \frac{(z - z_0)^{n+1}}{(z - z_0)^n} = z - z_0
Next, we evaluate the limit. Since the expression z - z_0 is independent of n, the limit is trivial:
L = \lim_{n \to \infty} \left| z - z_0 \right| = |z - z_0|
For the series to converge, we must have L < 1. This imposes the condition:
|z - z_0| < 1
This result confirms that the series converges inside an open disk of radius 1 centered at z_0.
The series is a geometric series with a first term a=1 and a common ratio r = (z - z_0). To find its sum, we first consider the finite partial sum, S_N:
S_N = \sum_{n=0}^N r^n = 1 + r + r^2 + \dots + r^N
We can find a closed-form expression for S_N. Multiply the entire equation by the common ratio r:
r S_N = r + r^2 + r^3 + \dots + r^{N+1}
Now, subtract this new equation from the original expression for S_N:
S_N - r S_N = (1 + r + r^2 + \dots + r^N) - (r + r^2 + \dots + r^{N+1})
Most terms on the right-hand side cancel out, leaving only the first term of S_N and the last term of rS_N:
S_N (1 - r) = 1 - r^{N+1}
Solving for S_N (assuming r \neq 1) gives the formula for the finite geometric sum:
S_N = \frac{1 - r^{N+1}}{1 - r}
The sum of the infinite series is defined as the limit of the partial sum S_N as N \to \infty.
\sum_{n=0}^\infty r^n = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{1 - r^{N+1}}{1 - r}
The value of this limit depends entirely on the behavior of the r^{N+1} term:
- If |r| < 1, then \lim_{N \to \infty} r^{N+1} = 0.
- If |r| > 1, the term |r|^{N+1} grows without bound, and the limit diverges.
- If |r| = 1 (and r \neq 1), the limit of r^{N+1} does not exist.
Therefore, the series converges if and only if |r| < 1. In this case, the sum becomes:
\sum_{n=0}^\infty r^n = \frac{1 - 0}{1 - r} = \frac{1}{1 - r}
By substituting our common ratio r = (z - z_0) back into the formula, we obtain the function to which the series converges:
f(z) = \sum_{n=0}^\infty (z - z_0)^n = \frac{1}{1 - (z - z_0)}
This equality holds precisely in the region where the series is convergent, which we found to be |z - z_0| < 1. The ratio test provided the domain of convergence, and the geometric sum formula provided the function that the series represents within that domain.