Taylor's Theorem and Series Expansion Techniques

The real problem of humanity is the following: We have Paleolithic emotions, medieval institutions and godlike technology. And it is terrifically dangerous, and it is now approaching a point of crisis overall, Edward O. Wilson.

Introduction

Definition. Complex sequence A sequence of complex numbers is a function $a: \mathbb{N} \to \mathbb{C}$. We usually denote it by $(a_n)_{n \in \mathbb{N}}$ or simply $(a_n)$, where $a_n := a(n)$. The value $a_1$ is called the first term of the sequence, $a_2$ the second term, and in general $a_n$ the n-th term of the sequence.

Definition. Convergent complex sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is said to converge to a complex number $L \in \mathbb{C}$ if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n \geq N$ one has $|a_n - L| < \varepsilon$. In this case we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$, and L is called the limit of the sequence $(a_n)_{n \in \mathbb{N}}$.

Definition. Cauchy sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is called a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n, m \geq N$ one has $|a_n - a_m| < \varepsilon$.

Definition. Series and partial sums.Let $(a_n)_{n \in \mathbb{N}}$ be a complex sequence. For each n $\in \mathbb{N}$, the finite sum $s_n := a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ is called the n-th partial sum of the (infinite) series $\sum_{k=1}^\infin a_k$ which we also denote simply by $\sum a_n$ when the index is clear from the context.

Definition. Convergent series. The series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $s \in \mathbb{C}$ if the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ defined by $s_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ converges to s, that is, $\lim_{n \to \infty} s_n = s$. In this case we write $s := \sum_{n=1}^\infin a_n$. If the sequence $(s_n)_{n \in \mathbb{N}}$ does not converge, we say that the series $\sum_{n=1}^{\infty} a_n$ diverges (or does not converge).

Definition. A complex power series centered at 0 in the variable z is a series of the form $a_0 + a_1z + a_2z^2 + \cdots = \sum_{n=0}^\infty a_n z^n$ with coefficients $a_i \in \mathbb{C}$

Definition. A complex power series centered at a complex number $a \in \mathbb{C} $ is an infinite series of the form: $\sum_{n=0}^\infty a_n (z - a)^n,$ where each $a_n \in \mathbb{C}$ is a coefficient, z is a complex variable, and $(z - a)^n$ is the nth power about the center.

Theorem. Given a power series $\sum_{n=0}^\infty a_n z^n$, there exists a unique value R, $0 \le R \le \infin$ (called the radius of convergence) such that:

1. For any z with |z| < R (inside the circle), the series $\sum_{n=0}^\infty a_n z^n$ converges absolutely (this is a “green light” zone).
2. For any z with |z| > R, the series diverges (this is a “red light” zone).

   On the Circle (|z| = R), this theorem gives no information. This is the yellow light zone —the series could converge or diverge.

Differentiability of Power Series. If $f(z) = \sum_{n=0}^{\infty} a_nz^n$ for |z| < R (R > 0), then f is analytic on B(0; R) and $f'(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$ for |z| < R.

Weierstrass M-test. Let $\{u_k(z)\}_{k=0}^\infty$ be a sequence of complex-valued functions defined on a set $\gamma^* \subseteq \mathbb{C}$. If there exists a sequence of non-negative real numbers $\{M_k\}_{k=0}^\infty$ such that:

1. Bounding Condition: $|u_k(z)| \leq M_k$ for all $z \in \gamma^*$ and all $k \in \mathbb{N}$
2. Convergence of Bound: The series $\sum_{k=0}^\infty M_k$ converges

Then, the original series $\sum_{k=0}^\infty u_k(z)$ converges uniformly on $\gamma^*$.

Coefficients of power series. Let f(z) = $\sum_{k=0}^\infty c_kz^k$ where this power series has radius of convergence R > 0. Then,the n-th coefficient of a power series $c_n$ can be extracted using the integral formula, $c_n = \frac{1}{2\pi i} \int_{C_r} \frac{f(z)}{z^{n+1}}dz, 0 \le r \lt R, n \ge 0$ where $C_r$ is a circle of radius r centered at 0 and oriented positively.

Taylor’s Theorem

Taylor’s Theorem. If f is analytic on an open disk B(a; R) (a disk of radius R centered at a), then f(z) can be represented exactly by a unique power series within that disk: $f(z) = \sum_{n=0}^{\infty}a_n (z - a)^n, \forall z \in B(a; R)$

This theorem bridges the gap between differentiability and power series. It guarantees that if a function behaves well (it is analytic) in a disk, it must also be infinitely differentiable and expressed or representable by a power series (an infinite polynomial) within that disk.

Furthermore, there exist unique constants $a_n = \frac{f^{(n)}(a)}{n!} = \frac{1}{2\pi i}\int_{C_r} \frac{f(w)}{(w-a)^{n+1}}dw$ where $C_r$ is a circle of radius r < R centered at a and oriented in the counterclockwise direction (positively oriented).

As we have demonstrated earlier, if a function $f$ is analytic, it is automatically infinitely differentiable, i.e., the expression $f^{(n)}(a)$ (the $n$-th derivative at $a$) makes sense. The Cauchy Integral Formula for Derivatives gives us a way to calculate $f^{(n)}(a)$ using an integral: $f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{C_r} \frac{f(w)}{(w-a)^{n+1}} dw$

Proof.

Fix a point z inside the disk B(a; R). We choose a radius r such that z sits inside the circle $C_r$, but $C_r$ is still inside the main disk, |z - a| < r < R.

According to Cauchy’s Integral Formula, the value of the function at z is determined by its values on the boundary circle, $C_r: f(z) = \frac{1}{2\pi i}\int_{C_r} \frac{f(w)}{w-z}dw$

We use an algebraic trick. $\frac{1}{w-z} = \frac{1}{(w-a) - (z-a)}$.

Factor out (w -a) from the denominator: $\frac{1}{w-z} = \frac{1}{w-a}\cdot \left[\frac{1}{1-\frac{z-a}{w-a}}\right]$

Remember the geometric series formula: $\frac{1}{1-q} = \sum_{n=0}^{\infty} q^n$, which is valid if $|q| < 1$. In our case, $q = \frac{z-a}{w-a}$. Because z is inside the circle and w is on the circle, $|z - a| < |w -a| = r, \forall w \in C_r, |\frac{z-a}{w-a}| \lt 1$.

We can now expand the bracketed term: $\frac{1}{w-z} = \frac{1}{w-a} \sum_{n=0}^{\infty} \frac{(z-a)^n}{(w-a)^n}$

Substitute this series back into our integral $f(z) = \frac{1}{2\pi i}\int_{C_r} \frac{f(w)}{w-z}dw = \frac{1}{2\pi i}\int_{C_r} \left( f(w)\cdot \frac{1}{w-a} \cdot \sum_{n=0}^{\infty} \frac{(z-a)^n}{(w-a)^n} \right) dw = \frac{1}{2\pi i}\oint_{C_r} \left( f(w) \sum_{n=0}^{\infty} \frac{(z-a)^n}{(w-a)^{n+1}} \right) dw$

$|w -a| = r, \forall w \in C_r$, Since f is analytic on an open disk B(a; R), it is continuous on the compact circle $C_r$, so it is bounded by some maximum value M, $\left| \frac{f(w)}{(w-a)^{n+1}} (z-a)^n \right| \le \frac{M}{r^{n+1}} |z-a|^n = \frac{M}{r} \left( \frac{|z-a|}{r} \right)^n$. Let’s define $M_n:= \frac{M}{r}(\frac{|z-a|}{r})^n$.

Since $\frac{|z-a|}{r} < 1$ (|z - a| < r), the sum of these bounds $\sum_{n=0}^{\infty} M_n$ is a convergent geometric series.

Lemma 1. Let $\gamma$ be a path, let $u_k(z)$ be a sequence of continuous function on $\gamma*$. If we can find a set of positive constants $M_k$ such that:

1. Bounding Condition. $|u_k(z)| \le M_k, \forall z \in \gamma*$ (each term is “dominated” by a constant).
2. Convergence of Bound. $\sum_{k=0}^\infin M_k$ converges.

Then, $\sum_{k=0}^\infin \int_\gamma u_k(z)dz$ converges uniformly to a continuous function U(z), and we can legally swap the integral and the sum: $\sum_{k=0}^\infin \int_\gamma u_k(z)dz = \int_\gamma \sum_{k=0}^\infin u_k(z)dz = \int_\gamma U(z)dz$.

By lemma 1, because the series is dominated by a convergent geometric series, it converges uniformly. Therefore, we are allowed to swap the integral and the sum, $f(z) = \sum_{n=0}^\infin \left[ \frac{1}{2\pi i} \int_{C_r} \frac{f(w)}{(w-a)^{n+1}}dw \right] (z-a)^n =[\text{By the Cauchy Integral Formula for Derivatives}] \sum_{n=0}^\infin \frac{f^{(n)}(a)}{n!} (z-a)^n$

Examples

• Determine the power series of f(z) = z⁶sin(3z).

This problem illustrates a crucial strategy in Complex Analysis: don’t compute derivatives if you don’t have to or why working less may make you more productive.

Using Taylor’s theorem, we can derive the power series for this function. It tells us that coefficients are $a_n = \frac{f^{(n)}(a)}{n!}$, calculating the 7th or 8th derivative of $z^6\sin(3z)$ using the Product Rule would be incredibly tedious and error-prone.

Instead of computing the derivatives directly, we derive the series for $\sin(3z)$ first, which simplifies the process.

The standard Maclaurin series for sin(w) is: $\sin(w) = w - \frac{w^3}{3!} + \frac{w^5}{5!} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^n w^{2n+1}}{(2n+1)!}$. It follows from the Taylor expansion at 0, using $\sin ^{(k)}(0)$ which alternates between 0 for even k and $\pm 1$ for odd k, giving only odd powers. The radius of convergence is infinite, so the series converges for all real or complex w ($R = \infty$).

We substitute the argument of our sine function, 3z, into the place of w in the series, $sin(3z) = \sum_{n=0}^{\infty} \frac{(-1)^n (3z)^{2n+1}}{(2n+1)!}, \forall z \in \mathbb{C}$ (the radius of converge is ∞).

Multiplying this by $z^6$ allows us to construct the series for f(z): $f(z) = z^6\sum_{n=0}^{\infty} \frac{(-1)^n (3z)^{2n+1}}{(2n+1)!}$

Because $z^6$ does not depend on the index n, we can bring it inside the summation. We then combine the exponents of $z$ using the rule $z^a \cdot z^b = z^{a+b}, z^6 \cdot z^{2n+1} = z^{(2n+1) + 6} = z^{2n+7}$

$f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n 3^{2n+1}z^{2n+7}}{(2n+1)!}$

• Let’s consider the principal branch of the complex logarithm. The domain is $\mathbb{C} \setminus (-\infin, 0]$ (the entire complex plane and remove the origin 0 and the negative real axis). Define f on $\mathbb{C} \setminus (-\infin, 0]$ as $ f(z) = Log(z) = \log |z| + i\theta$ where $\theta \in arg(z), -\pi \lt \theta \lt \pi$.

  This “cut” is necessary to make the function continuous and analytic.

Since $f(z)$ is analytic on this domain, Taylor’s Theorem guarantees a series exists. The radius of convergence R is the distance from the center (a = 1) to the nearest singularity (problem point). The nearest singularity is at z = 0 (where $\log(z)$ is undefined), the distance from 1 to 0 is 1. Therefore, the series is valid for $|z - 1| < 1$.

Instead of calculating derivatives of $\text{Log}(z)$ directly (which gets messy), we use a clever shortcut:

1. We know the derivative of $\text{Log}(z)$ is $\frac{1}{z}$.
2. We can easily write the series for $\frac{1}{z}$ using the Geometric Series Formula: $\frac{1}{1-q} = \sum_{n=0}^{\infty} q^n$. $\frac{1}{z} = \frac{1}{1+(z-1)} = \frac{1}{1-(-(z-1))} = \sum_{n=0}^{\infty} (-1)^n(z-1)^n, \forall z \in \mathbb{B}(1; 1)$.

   Here, our “ratio” q is -(z - 1). Since |z - 1| < 1, the series converges.

3. Log(z) is analytic on the domain $\mathbb{C} \setminus (-\infin, 0]$ and $\frac{d}{dz}(log(z)) = \frac{1}{z} = \sum_{n=0}^{\infty} (-1)^n(z-1)^n, \forall z \in \mathbb{B}(1; 1)$
4. Let us assume the Taylor series for our function exists with unknown coefficients $c_n, \text{Log}(z) = \sum_{n=0}^{\infty} c_n (z-1)^n = c_0 + c_1(z-1) + c_2(z-1)^2 + \dots, \forall z \in \mathbb{B}(1; 1)$. Now, differentiate this series term-by-term: $\frac{d}{dz}(\text{Log}(z)) = 0 + 1\cdot c_1 + 2\cdot c_2(z-1) + 3\cdot c_3(z-1)^2 + \dots = \sum_{n=1}^{\infty} n c_n (z-1)^{n-1}, \forall z \in \mathbb{B}(1; 1)$.
5. Now we have two expressions for the derivative. They must be equal. $nc_n = (-1)^{n-1}, \forall n \ge 1$. Solving for $c_n, c_n = \frac{(-1)^{n-1}}{n} \quad \text{for } n \ge 1$. The formula above works for $n \ge 1$. We still need the constant term $c_0$. We can plug the center point z = 1 into the original definition: f(1) = Log(1) = 0. $f(1) = c_0 + c_1(0) + c_2(0) \dots = c_0$. Therefore, $c_0 = 0$.
6. Substituting our coefficients back into the series representation: $f(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(z-1)^n$, $\forall z \in \mathbb{B}(1; 1)$. This specific series is famously known as the Mercator Series. It comes from Nicholas Mercator (1620–1687), who published this expansion in his work Logarithmotechnia (1668). The series converges for |z - 1| < 1, i.e. inside the open disk $\mathbb{B}(1;1)$.

• Consider the complex function $\frac{1}{1+z^2}$. Before we find the series, we must know where it lives (the radius of convergence). A power series is valid only within a disk that avoids singularities or “explosions”, points where the function ceases to be analytic (blows up, becomes multivalued, or undefined).

Find the trouble spots: The function is undefined when the denominator is zero, $1 + z^2 = 0 \implies z^2 = -1 \implies z = \pm i$. We are expanding around the center a = 0. The distance from 0 to the nearest singularities (i and -i) is 1. Therefore, the series will converge strictly inside the unit disk: $B(0; 1)$, or |z| < 1.

Calculating derivatives of this function gets very ugly really fast. Instead, we use the Geometric Series formula again, $\frac{1}{1-q} = \sum_{n=0}^{\infty} q^n$ (valid for |q| < 1). We can rewrite the addition in the denominator as a subtraction of a negative: $f(z) = \frac{1}{1 - (-z^2)}$

Our “q” is: $-z^2$ and we need $|-z^2| < 1$, which simplifies to $|z| < 1$. This confirms our previous radius of convergence analysis. Substitute $q = -z^2$ into the summation: $f(z) = \sum_{n=0}^{\infty} (-z^2)^n$.

We can separate the negative sign to clean up the expression, $(-z^2)^n = (-1 \cdot z^2)^n = (-1)^n \cdot (z^2)^n = (-1)^n z^{2n}, f(z) = \sum_{n=0}^{\infty} (-1)^n z^{2n} = 1 - z^2 + z^4 - z^6 + \cdots$ (the series is an alternating sum of even powers).

This function is the derivative of another famous function: $\arctan(z)$. Because we now have the series for $\frac{1}{1+z^2}$, we can integrate it term-by-term to find the series for $\arctan(z)$, $\arctan(z) = \int \left( 1 - z^2 + z^4 - \dots \right) dz = z - \frac{z^3}{3} + \frac{z^5}{5} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{2n+1}$.