Editing Taylor's theorem (section)

== Relationship to analyticity ==

=== Taylor expansions of real analytic functions ===

Let ''I'' ⊂ '''R''' be an [[open interval]]. By definition, a function ''f'' : ''I'' → '''R''' is [[analytic function|real analytic]] if it is locally defined by a convergent [[power series]]. This means that for every ''a''&nbsp;∈&nbsp;''I'' there exists some ''r''&nbsp;>&nbsp;0 and a sequence of coefficients ''c<sub>k</sub>''&nbsp;∈&nbsp;'''R''' such that {{nowrap|(''a'' − ''r'', ''a'' + ''r'') ⊂ ''I''}} and

<math display="block"> f(x) = \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots, \qquad |x-a|<r. </math>

In general, the [[power series#Radius of convergence|radius of convergence]] of a power series can be computed from the [[Cauchy–Hadamard theorem|Cauchy–Hadamard formula]]

<math display="block"> \frac{1}{R} = \limsup_{k\to\infty}|c_k|^\frac{1}{k}. </math>

This result is based on comparison with a [[geometric series]], and the same method shows that if the power series based on ''a'' converges for some ''b'' ∈ '''R''', it must converge [[uniform convergence|uniformly]] on the [[closed interval]] <math display="inline">[a-r_b,a+r_b]</math>, where <math display="inline">r_b=\left\vert b-a \right\vert</math>. Here only the convergence of the power series is considered, and it might well be that {{nowrap|(''a'' − ''R'',''a'' + ''R'')}} extends beyond the domain ''I'' of ''f''.

The Taylor polynomials of the real analytic function ''f'' at ''a'' are simply the finite truncations

<math display="block"> P_k(x) = \sum_{j=0}^k c_j(x-a)^j, \qquad c_j = \frac{f^{(j)}(a)}{j!}</math>

of its locally defining power series, and the corresponding remainder terms are locally given by the analytic functions

<math display="block"> R_k(x) = \sum_{j=k+1}^\infty c_j(x-a)^j = (x-a)^k h_k(x), \qquad |x-a|<r. </math>

Here the functions

<math display="block">\begin{align}
& h_k:(a-r,a+r)\to \R \\[1ex]
& h_k(x) = (x-a)\sum_{j=0}^\infty c_{k+1+j} \left(x - a\right)^j
\end{align}</math>

are also analytic, since their defining power series have the same radius of convergence as the original series. Assuming that {{nowrap|[''a'' − ''r'', ''a'' + ''r'']}} ⊂ ''I'' and ''r''&nbsp;<&nbsp;''R'', all these series converge uniformly on {{nowrap|(''a'' − ''r'', ''a'' + ''r'')}}. Naturally, in the case of analytic functions one can estimate the remainder term <math display="inline">R_k(x)</math> by the tail of the sequence of the derivatives ''f′''(''a'') at the center of the expansion, but using [[complex analysis]] also another possibility arises, which is described [[Taylor's theorem#Relationship to analyticity##Taylor's theorem in complex analysis|below]].

=== Taylor's theorem and convergence of Taylor series ===

The Taylor series of ''f'' will converge in some interval in which all its derivatives are bounded and do not grow too fast as ''k'' goes to infinity. (However, even if the Taylor series converges, it might not converge to ''f'', as explained below; ''f'' is then said to be non-[[analytic function|analytic]].)

One might think of the Taylor series

<math display="block"> f(x) \approx \sum_{k=0}^\infty c_k(x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots </math>

of an infinitely many times differentiable function ''f'' : '''R''' → '''R''' as its "infinite order Taylor polynomial" at ''a''. Now the [[Taylor's theorem#Estimates for the remainder|estimates for the remainder]] imply that if, for any ''r'', the derivatives of ''f'' are known to be bounded over (''a''&nbsp;−&nbsp;''r'', ''a''&nbsp;+&nbsp;''r''), then for any order ''k'' and for any ''r''&nbsp;>&nbsp;0 there exists a constant {{nowrap|''M<sub>k,r</sub>'' > 0}} such that

{{NumBlk|:|<math> |R_k(x)| \leq M_{k,r} \frac{|x-a|^{k+1}}{(k+1)!} </math>|{{EquationRef|★★}}}}

for every ''x''&nbsp;∈&nbsp;(''a''&nbsp;−&nbsp;''r'',''a''&nbsp;+&nbsp;''r''). Sometimes the constants {{nowrap|''M<sub>k,r</sub>''}} can be chosen in such way that {{nowrap|''M<sub>k,r</sub>''}} is bounded above, for fixed ''r'' and all ''k''. Then the Taylor series of ''f'' [[uniform convergence|converges uniformly]] to some analytic function

<math display="block">\begin{align}
& T_f:(a-r,a+r)\to\R \\
& T_f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} \left(x-a\right)^k
\end{align}</math>

(One also gets convergence even if {{nowrap|''M<sub>k,r</sub>''}} is not bounded above as long as it grows slowly enough.)

The limit function {{nowrap|''T<sub>f</sub>''}} is by definition always analytic, but it is not necessarily equal to the original function ''f'', even if ''f'' is infinitely differentiable. In this case, we say ''f'' is a [[non-analytic smooth function]], for example a [[flat function]]:

<math display="block">\begin{align}
& f:\R \to \R \\
& f(x) = \begin{cases}
e^{-\frac{1}{x^2}} & x>0  \\
0 & x \leq 0 .
\end{cases}
\end{align}</math>

Using the [[chain rule]] repeatedly by [[mathematical induction]], one shows that for any order&nbsp;''k'',

<math display="block"> f^{(k)}(x) = \begin{cases}
\frac{p_k(x)}{x^{3k}}\cdot e^{-\frac{1}{x^2}}  & x>0 \\
0 & x \leq 0
\end{cases}</math>

for some polynomial ''p<sub>k</sub>'' of degree 2(''k'' − 1). The function <math>e^{-\frac{1}{x^2}}</math> tends to zero faster than any polynomial as <math display="inline">x \to 0</math>, so ''f'' is infinitely many times differentiable and {{nowrap|1=''f''{{i sup|(''k'')}}(0) = 0}} for every positive integer ''k''. The above results all hold in this case:

* The Taylor series of ''f'' converges uniformly to the zero function ''T<sub>f</sub>''(''x'')&nbsp;=&nbsp;0, which is analytic with all coefficients equal to zero.
* The function ''f'' is unequal to this Taylor series, and hence non-analytic.
* For any order ''k''&nbsp;∈&nbsp;'''N''' and radius ''r''&nbsp;>&nbsp;0 there exists ''M<sub>k,r</sub>''&nbsp;>&nbsp;0 satisfying the remainder bound&nbsp;({{EquationNote|★★}}) above.
However, as ''k'' increases for fixed ''r'', the value of ''M<sub>k,r</sub>''  grows more quickly than ''r<sup>k</sup>'', and the error does not go to zero''.''

=== 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>

=== Example ===

[[File:Function with two poles.png|thumb|right|Complex plot of <math display="inline">f(z)=\frac{1}{1+z^2}</math>. Modulus is shown by elevation and argument by coloring: cyan&nbsp;=&nbsp;<math display="inline">0</math>, blue&nbsp;=&nbsp;<math display="inline">\frac{\pi}{3}</math>, violet&nbsp;=&nbsp;<math display="inline">\frac{2\pi}{3}</math>, red&nbsp;=&nbsp;<math>\pi</math>, yellow&nbsp;=&nbsp;<math display="inline">\frac{4\pi}{3}</math>, green&nbsp;=&nbsp;<math display="inline">\frac{5\pi}{3}</math>.]]
The function

<math display="block">\begin{align}
& f : \R \to \R \\
& f(x) = \frac{1}{1+x^2}
\end{align}</math>

is [[analytic function|real analytic]], that is, locally determined by its Taylor series. This function was plotted [[Taylor's theorem#Motivation|above]] to illustrate the fact that some elementary functions cannot be approximated by Taylor polynomials in neighborhoods of the center of expansion which are too large. This kind of behavior is easily understood in the framework of complex analysis. Namely, the function ''f'' extends into a [[meromorphic function]]

<math display="block">\begin{align}
& f:\Complex \cup \{\infty\} \to \Complex \cup \{\infty\} \\
& f(z) = \frac{1}{1+z^2}
\end{align}</math>

on the compactified complex plane. It has simple poles at <math display="inline">z=i</math> and <math display="inline">z=-i</math>, and it is analytic elsewhere. Now its Taylor series centered at ''z''<sub>0</sub> converges on any disc ''B''(''z''<sub>0</sub>, ''r'') with ''r'' < |''z''&nbsp;−&nbsp;''z''<sub>0</sub>|, where the same Taylor series converges at ''z''&nbsp;∈&nbsp;'''C'''. Therefore, Taylor series of ''f'' centered at 0 converges on ''B''(0, 1) and it does not converge for any ''z'' ∈ '''C''' with |''z''|&nbsp;>&nbsp;1 due to the poles at ''i'' and −''i''. For the same reason the Taylor series of ''f'' centered at 1 converges on <math display="inline">B(1, \sqrt{2})</math> and does not converge for any ''z''&nbsp;∈&nbsp;'''C''' with <math display="inline">\left\vert z-1 \right\vert>\sqrt{2}</math>.