Question 6.22: Prove Taylor’s theorem: If f(z) is analytic inside a circle ...
Prove Taylor’s theorem: If f(z) is analytic inside a circle C with center at a, then for all z inside C,
f(z)=f(a)+f^{\prime}(a)(z-a)+\frac{f^{\prime \prime}(a)}{2 !}(z-a)^2+\frac{f^{\prime \prime \prime}(a)}{3 !}(z-a)^3+\cdotsLet z be any point inside C. Construct a circle C_1 with center at a and enclosing z (see Fig. 6-4). Then, by Cauchy’s integral formula,
f(z)=\frac{1}{2 \pi i} \oint_{C_1} \frac{f(w)}{w-z} d w (1)
We have
\begin{aligned}\frac{1}{w-z}= & \frac{1}{(w-a)-(z-a)}=\frac{1}{w-a}\left\{\frac{1}{1-(z-a) /(w-a)}\right\} \\= & \frac{1}{w-a}\left\{1+\left(\frac{z-a}{w-a}\right)+\left(\frac{z-a}{w-a}\right)^2+\cdots+\left(\frac{z-a}{w-a}\right)^{n-1}\right. \\& \left.+\left(\frac{z-a}{w-a}\right)^n \frac{1}{1-(z-a) /(w-a)}\right\}\end{aligned}or
\frac{1}{w-z}=\frac{1}{w-a}+\frac{z-a}{(w-a)^2}+\frac{(z-a)^2}{(w-a)^3}+\cdots+\frac{(z-a)^{n-1}}{(w-a)^n}+\left(\frac{z-a}{w-a}\right)^n \frac{1}{w-z} (2)
Multiplying both sides of (2) by f(w) and using (1), we have
f(z)=\frac{1}{2 \pi i} \oint_{C_1} \frac{f(w)}{w-a} d w+\frac{z-a}{2 \pi i} \oint_{C_1} \frac{f(w)}{(w-a)^2} d w+\cdots+\frac{(z-a)^{n-1}}{2 \pi i} \oint_{C_1} \frac{f(w)}{(w-a)^n} d w+U_n (3)
where
U_n=\frac{1}{2 \pi i} \oint_{C_1}\left(\frac{z-a}{w-a}\right)^n \frac{f(w)}{w-z} d wUsing Cauchy’s integral formulas
f^{(n)}(a)=\frac{n !}{2 \pi i} \oint_{C_1} \frac{f(w)}{(w-a)^{n+1}} d w \quad n=0,1,2,3, \ldots(3) becomes
f(z)=f(a)+f^{\prime}(a)(z-a)+\frac{f^{\prime \prime}(a)}{2 !}(z-a)^2+\cdots+\frac{f^{(n-1)}(a)}{(n-1) !}(z-a)^{n-1}+U_nIf we can now show that \lim _{n \rightarrow \infty} U_n=0, we will have proved the required result. To do this, we note that since w is on C_1,
\left|\frac{z-a}{w-a}\right|=\gamma<1where \gamma is a constant. Also, we have |f(w)|<M where $M$ is a constant, and
|w-z|=|(w-a)-(z-a)| \geq r_1-|z-a|where r_1 is the radius of C_1. Hence, from Property (e), Page 112, we have
\begin{aligned}\left|U_n\right| & =\frac{1}{2 \pi}\left|\oint_{C_1}\left(\frac{z-a}{w-a}\right)^n \frac{j(w)}{w-z} d w\right| \\& \leq \frac{1}{2 \pi} \frac{\gamma^n M}{r_1-|z-a|} \cdot 2 \pi r_1=\frac{\gamma^n M r_1}{r_1-|z-a|}\end{aligned}and we see that \lim _{n \rightarrow \infty} U_n=0, completing the proof.
