Editing Taylor's theorem (section)

=== Taylor's theorem in complex analysis ===

Taylor's theorem generalizes to functions ''f'' : '''C''' → '''C''' which are [[complex differentiable]] in an open subset ''U''&nbsp;⊂&nbsp;'''C''' of the [[complex plane]]. However, its usefulness is dwarfed by other general theorems in [[complex analysis]]. Namely, stronger versions of related results can be deduced for [[complex differentiable]] functions ''f''&nbsp;:&nbsp;''U''&nbsp;→&nbsp;'''C''' using [[Cauchy's integral formula]] as follows.

Let ''r''&nbsp;>&nbsp;0 such that the [[closed disk]] ''B''(''z'',&nbsp;''r'')&nbsp;∪&nbsp;''S''(''z'',&nbsp;''r'') is contained in ''U''. Then Cauchy's integral formula with a positive parametrization {{nowrap|1=''γ''(''t'') = ''z'' + ''re<sup>it</sup>''}} of the circle ''S''(''z'', ''r'') with <math>t \in [0,2 \pi]</math>  gives

<math display="block">f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}\,dw, \quad f'(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^2} \, dw, \quad \ldots, \quad f^{(k)}(z) = \frac{k!}{2\pi i}\int_\gamma \frac{f(w)}{(w-z)^{k+1}} \, dw.</math>

Here all the integrands are continuous on the [[circle]] ''S''(''z'',&nbsp;''r''), which justifies differentiation under the integral sign. In particular, if ''f'' is once [[complex differentiable]]  on the open set ''U'', then it is actually infinitely many times [[complex differentiable]] on ''U''. One also obtains [[Cauchy's estimate]]<ref>{{harvnb|Rudin|1987|loc=§10.26}}</ref>

<math display="block"> |f^{(k)}(z)| \leq \frac{k!}{2\pi}\int_\gamma \frac{M_r}{|w-z|^{k+1}} \, dw = \frac{k!M_r}{r^k}, \quad M_r = \max_{|w-c|=r}|f(w)| </math>

for any ''z''&nbsp;∈&nbsp;''U'' and ''r''&nbsp;>&nbsp;0 such that ''B''(''z'',&nbsp;''r'')&nbsp;∪&nbsp;''S''(''c'',&nbsp;''r'')&nbsp;⊂&nbsp;''U''. The estimate implies that the [[complex number|complex]] [[Taylor series]]

<math display="block"> T_f(z) =  \sum_{k=0}^\infty \frac{f^{(k)}(c)}{k!}(z-c)^k </math>

of ''f'' converges uniformly on any [[open disk]] <math display="inline">B(c,r) \subset U</math> with <math display="inline">S(c,r) \subset U</math> into some function ''T<sub>f</sub>''. Furthermore, using the [[contour integral]] formulas for the derivatives ''f''{{i sup|(''k'')}}(''c''),

<math display="block">\begin{align} 
T_f(z) &= \sum_{k=0}^\infty  \frac{(z-c)^k}{2\pi i}\int_\gamma \frac{f(w)}{(w-c)^{k+1}} \, dw \\
&=  \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-c} \sum_{k=0}^\infty  \left(\frac{z-c}{w-c}\right)^k \, dw \\
&= \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-c}\left( \frac{1}{1-\frac{z-c}{w-c}} \right) \, dw \\
&= \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-z} \, dw \\
&= f(z),
\end{align}</math>

so any [[complex derivative|complex differentiable]] function ''f'' in an open set ''U''&nbsp;⊂&nbsp;'''C''' is in fact [[complex analytic]]. All that is said for real analytic functions [[Taylor's theorem#Relationship to analyticity##Taylor expansions of analytic functions|here]] holds also for complex analytic functions with the open interval ''I'' replaced by an open subset ''U''&nbsp;∈&nbsp;'''C''' and ''a''-centered intervals (''a''&nbsp;−&nbsp;''r'',&nbsp;''a''&nbsp;+&nbsp;''r'') replaced by ''c''-centered disks ''B''(''c'',&nbsp;''r''). In particular, the Taylor expansion holds in the form

<math display="block"> f(z) = P_k(z) + R_k(z), \quad P_k(z) = \sum_{j=0}^k \frac{f^{(j)}(c)}{j!}(z-c)^j, </math>

where the remainder term ''R<sub>k</sub>'' is complex analytic. Methods of complex analysis provide some powerful results regarding Taylor expansions. For example, using Cauchy's integral formula for any positively oriented [[Jordan curve]] <math display="inline">\gamma</math> which parametrizes the boundary <math display="inline">\partial W \subset U</math> of a region <math display="inline">W \subset U</math>, one obtains expressions for the derivatives {{nowrap|''f''{{i sup|(''j'')}}(''c'')}} as above, and modifying slightly the computation for {{nowrap|1=''T<sub>f</sub>''(''z'') = ''f''(''z'')}}, one arrives at the exact formula

<math display="block"> R_k(z) = \sum_{j=k+1}^\infty  \frac{(z-c)^j}{2\pi i} \int_\gamma \frac{f(w)}{(w-c)^{j+1}} \, dw = \frac{(z-c)^{k+1}}{2\pi i} \int_\gamma \frac{f(w) \, dw}{(w-c)^{k+1}(w-z)} , \qquad z\in W. </math>

The important feature here is that the quality of the approximation by a Taylor polynomial on the region <math display="inline">W \subset U</math> is dominated by the values of the function ''f'' itself on the boundary <math display="inline">\partial W \subset U</math>. Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates

<math display="block"> |R_k(z)|
\leq \sum_{j=k+1}^\infty \frac{M_r |z-c|^j}{r^j}
= \frac{M_r}{r^{k+1}} \frac{|z-c|^{k+1}}{1-\frac{|z-c|}{r}}
\leq \frac{M_r \beta^{k+1}}{1-\beta}, \qquad \frac{|z-c|}{r} \leq \beta < 1. </math>