## Proof of Tayor's theorem for analytic functions

"Figure 1: The circle of convergence C in the complex w plane"

By Cauchy's integral formula

$f(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{w-z}}\;dw$ .

Adding and subtracting the value $a$ in the denominator, and rewriting, we have

$f(z)={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{w-a}}\left({\frac {1}{1-{\frac {z-a}{w-a}}}}\right)\;dw$

We may expand the factor into a geometric series, provided that $|z-a|<|w-a|$ meaning that points of $a$
and $z$ lie inside $C$ and points of $w$ lie on $C$ and that $C$ is a disc of radius $|w-a|$ called the *circle of convergence* of the Taylor's series. See Figure 1.

$f(z)={\frac {1}{2\pi i}}\sum _{n=0}^{N-1}\left[\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw\right](z-a)^{n}+R_{N}$ .

From Cauchy's integral formulas we recognize

${\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-a)^{n+1}}}\;dw={\frac {f^{(n)}(a)}{n!}}.$
The only thing that remains is to show that the remainder $R_{N}$ vanishes as $N\rightarrow \infty$.

$R_{N}={\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw$ .

$|R_{N}|=\left|{\frac {1}{2\pi i}}\oint _{C}{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\;dw\right|\leq {\frac {1}{2\pi }}\oint _{C}\left|{\frac {f(w)}{(w-z)}}\left({\frac {z-a}{w-a}}\right)^{N}\right|\;dw$

We note that

$|w-z|=|w-a-(z-a)|\geq |w-a|-|z-a|$

$|z-a|<|w-a|.$
Thus we may define

$\gamma ={\frac {z-a}{w-a}}<1$

which yields the estimate

$|R_{n}|<{\frac {1}{2\pi }}M(2\pi |w-a|){\frac {\gamma ^{N}}{|w-a|-|z-a|}}={\frac {M\gamma ^{N}|w-a|}{|w-a|-|z-a|}}\rightarrow 0\qquad {\mbox{as}}\qquad N\rightarrow \infty$ .

Here the constant factor $M$ follows from the Maximum Modulus Theorem.

Named for Brook Taylor (1685–1731), English mathematician.