Editing Taylor's theorem (section)

=== 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''.''