---
title: "State and prove Taylor's series"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:taylor-series-prove"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-prove
site: BCT Engineering
---

# State and prove Taylor's series

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2075 Baisakh, 5 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-prove
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

State and prove Taylor's series for function of complex variable.

## Short answer

Taylor’s theorem: If f is analytic inside and on |z−a|=R, then f(z)=Σ f⁽ⁿ⁾(a)(z−a)ⁿ/n! for |z−a|<R. Proof: Use Cauchy integral formula, expand 1/(ζ−z) as geometric series in (z−a)/(ζ−a), integrate termwise, and apply the derivative formula.

## Answer

**Statement (Taylor's theorem).** Let f(z) be analytic inside and on a circle C: |z − a| = R. Then for every z inside C,

f(z) = f(a) + f′(a)(z − a) + f″(a)(z − a)²/2! + … = Σ_{n=0}^{∞} f⁽ⁿ⁾(a)(z − a)ⁿ/n!

#### Proof
Take any point z inside C and write |z − a| = r < R. By Cauchy's integral formula,

f(z) = (1/2πi) ∮_C f(ζ)/(ζ − z) dζ.  ... (1)

Write ζ − z = (ζ − a) − (z − a) = (ζ − a)[1 − (z − a)/(ζ − a)].

For ζ on C, |ζ − a| = R and |z − a| = r < R, so |(z − a)/(ζ − a)| = r/R < 1. The geometric series applies:

```text
1/(ζ − z) = 1/(ζ − a) · 1/[1 − (z − a)/(ζ − a)]
          = Σ_{n=0}^{∞} (z − a)ⁿ / (ζ − a)ⁿ⁺¹
```

By the Weierstrass M-test the series converges uniformly on C (terms bounded by (1/R)(r/R)ⁿ), and f(ζ) is bounded on C. So we can integrate term by term in (1):

f(z) = Σ_{n=0}^{∞} (z − a)ⁿ · (1/2πi) ∮_C f(ζ)/(ζ − a)ⁿ⁺¹ dζ.

By Cauchy's integral formula for derivatives,

(1/2πi) ∮_C f(ζ)/(ζ − a)ⁿ⁺¹ dζ = f⁽ⁿ⁾(a)/n!.

Therefore

f(z) = Σ_{n=0}^{∞} [f⁽ⁿ⁾(a)/n!] (z − a)ⁿ. ∎

**Remarks.**
- The series converges in the largest circle |z − a| < R centred at *a* inside which f is analytic (R is the distance to the nearest singularity).
- When a = 0, it is called the Maclaurin series.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The proof starts from which result?
   A. Cauchy's integral formula for f(z)
   B. Laurent series
   C. Residue theorem
   D. Green's theorem
   Answer: A. f(z) = (1/2πi)∮ f(ζ)/(ζ − z) dζ. (Key point: Start)

2. Expanding 1/(ζ − z) as a geometric series works because:
   A. ζ = z
   B. f is constant
   C. |z − a|/|ζ − a| = r/R < 1
   D. |z − a| > |ζ − a|
   Answer: C. ratio < 1 for ζ on C. (Key point: Geometric series)

3. Integration term by term is justified by:
   A. f being real
   B. Uniform convergence (Weierstrass M-test)
   C. Taking R → 0
   D. Differentiability of the series
   Answer: B. Terms are bounded by (1/R)(r/R)^n. (Key point: Term by term)

4. The coefficient of (z − a)^n is identified using:
   A. Cauchy–Riemann equations
   B. Cauchy's formula for derivatives, f^n(a)/n!
   C. Green's theorem
   D. The residue at infinity
   Answer: B. (1/2πi)∮ f(ζ)/(ζ−a)^{n+1} dζ = f^n(a)/n!. (Key point: Coefficient)

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