Taylor’s Series for Analytic Functions

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2017)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~20 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

Taylor’s theorem in the complex plane is one of the deepest results of complex analysis: it says that every analytic function is a power series in disguise. UPSC 2017 asked for an explicit Taylor expansion of an analytic function about a given point together with the radius of convergence — a standard but non-trivial 15-mark task. Understanding Taylor series also prepares you for Laurent series (singular points, residues) which appear far more in the exam.

Minimum Theory

Taylor’s Theorem for Analytic Functions

Theorem. Let $f$ be analytic in the disk $|z - z_0| < R$. Then $f$ has a convergent power series representation (Taylor series) in that disk:

$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n, \quad |z - z_0| < R,$

where the coefficients are uniquely determined by

$a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\,d\zeta,$

with $C$ any positively-oriented simple closed curve enclosing $z_0$ inside $|z - z_0| < R$.

The radius of convergence of the Taylor series equals the distance from $z_0$ to the nearest singularity of $f$ in $\mathbb{C}$ (or $+\infty$ if $f$ is entire).

Key Features

• Uniqueness. The Taylor series of $f$ at $z_0$ is unique; any other convergent power series expansion of $f$ at $z_0$ must be the Taylor series.
• Term-by-term differentiation. The Taylor series may be differentiated any number of times within $|z - z_0| < R$, yielding the Taylor series of $f'$.
• Radius = distance to nearest singularity. For $f(z) = 1/(1-z)$ expanded at $z_0 = 0$: the singularity is at $z = 1$, so $R = 1$.

Standard Taylor Series (memorise these)

Expanding at a Point $z_0 \neq 0$

Substitution method (fastest): Write $z = z_0 + w$ so $w = z - z_0$, expand as a series in $w$, then re-express in $(z - z_0)$.

Direct method (using Cauchy’s formula for coefficients): compute $a_n = f^{(n)}(z_0)/n!$ by differentiating $n$ times.

For most UPSC problems, the substitution method is faster and less error-prone.

Radius of Convergence from the Expansion

Once you have the Taylor series $\sum a_n(z-z_0)^n$, apply Hadamard or the ratio test as in P2-CX-06. Alternatively, simply state: $R =$ distance from $z_0$ to the nearest singularity of $f$.

Question Archetypes

expand-at-point (1 question; 2017)

Recognition Cues

• An analytic function and a centre point $z_0$ are given.
• Asked for the Taylor series and/or the radius of convergence.
• Function is typically a rational, exponential, or trigonometric function.

Solution Template

1. Identify the centre $z_0$ and locate the singularities of $f$.
2. State $R =$ distance from $z_0$ to nearest singularity.
3. Write $w = z - z_0$ and express $f$ in terms of $w$.
4. Expand using standard series (geometric, exponential, etc.) in powers of $w$.
5. Collect terms and write the series explicitly up to the degree requested (or in sigma notation).
6. Convert back to powers of $(z - z_0)$.
7. Confirm radius of convergence using Hadamard or the singularity-distance interpretation.

Worked Example

2017 Paper 2, 2017-P2-QCX (15 marks)

Expand $f(z) = \dfrac{1}{(z-1)(z-2)}$ as a Taylor series about $z_0 = 0$ and find the radius of convergence.

Step 1: Singularities and radius.

$f$ has simple poles at $z = 1$ and $z = 2$. The nearest singularity to $z_0 = 0$ is $z = 1$ (distance $1$). Therefore

$R = 1.$

The Taylor series converges in $|z| < 1$.

Step 2: Partial fractions.

$\frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}.$

Multiply through: $1 = A(z-2) + B(z-1)$.

Set $z = 1$: $1 = A(-1)$, so $A = -1$. Set $z = 2$: $1 = B(1)$, so $B = 1$.

$f(z) = \frac{-1}{z-1} + \frac{1}{z-2} = \frac{1}{1-z} - \frac{1}{2-z}.$

Step 3: Expand each term using the geometric series $\dfrac{1}{1-w} = \sum_{n=0}^\infty w^n$ for $|w| < 1$.

$\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n, \quad |z| < 1.$

$\frac{1}{2-z} = \frac{1}{2} \cdot \frac{1}{1 - z/2} = \frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}, \quad |z| < 2.$

Step 4: Combine.

$f(z) = \sum_{n=0}^{\infty} z^n - \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} = \sum_{n=0}^{\infty} \left(1 - \frac{1}{2^{n+1}}\right) z^n, \quad |z| < 1.$

Step 5: State the result.

$\boxed{f(z) = \sum_{n=0}^{\infty} \frac{2^{n+1} - 1}{2^{n+1}}\, z^n, \quad |z| < 1.}$

The radius of convergence is $R = 1$, determined by the singularity at $z = 1$.

Verification: The coefficient $a_n = 1 - 1/2^{n+1} = (2^{n+1}-1)/2^{n+1}$. Hadamard: $\limsup|a_n|^{1/n} \to 1$ (since $a_n \to 1$), confirming $R = 1$. $\blacksquare$

Common Traps

• Expanding $1/(2-z)$ without factoring out $1/2$ first: forgetting to factor gives an incorrect geometric series.
• Quoting $R = 2$ (distance to the farther singularity) instead of $R = 1$ (distance to the nearest).
• Using the Taylor series outside its radius: $\sum z^n$ diverges for $|z| \geq 1$.
• Mixing up signs in partial fractions: double-check by re-combining.
• For “prove Taylor’s theorem” questions: a full proof requires Cauchy’s integral formula for $f(z)$ and the geometric series expansion of $1/(\zeta - z)$; do not omit this.

Marks-Aware Writing

This is a 15-mark Section B question. UPSC expects:

1. Partial fractions (or other algebraic setup) — show all algebra.
2. Standard series identification — cite which standard series you are using.
3. Final series in sigma notation with explicit general term.
4. Radius of convergence — justify by singularity-distance or by Hadamard.

For “prove Taylor’s theorem” variants (which also appear), structure as: (a) state the theorem with hypotheses, (b) start from Cauchy’s integral formula, (c) expand $1/(\zeta - z)$ as a geometric series in $(z - z_0)/(\zeta - z_0)$, (d) justify term-by-term integration by uniform convergence, (e) identify coefficients.

Practice Set

Only one historical question on this atom (shown above).