State and prove Taylor's series

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:

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.