Editing Taylor's theorem (section)

== Taylor's theorem in one real variable ==

=== Statement of the theorem ===

The precise statement of the most basic version of Taylor's theorem is as follows:

{{math theorem
| name = Taylor's theorem<ref>{{ citation|first1=Angelo|last1=Genocchi|first2= Giuseppe|last2=Peano|title=Calcolo differenziale e principii di calcolo integrale|location=(N. 67, pp.&nbsp;XVII–XIX)|publisher=[[Fratelli Bocca |Fratelli Bocca ed.]]|year=1884}}</ref><ref>{{Citation | last1=Spivak | first1=Michael | author1-link=Michael Spivak | title=Calculus  | publisher=Publish or Perish | location=Houston, TX | edition=3rd | isbn=978-0-914098-89-8 | year=1994| page=383}}</ref><ref>{{springer|title=Taylor formula|id=p/t092300}}</ref>
| math_statement = Let ''k''&nbsp;≥&nbsp;1 be an [[integer]] and let the [[Function (mathematics)|function]] {{nowrap|''f'' : '''R''' → '''R'''}} be ''k'' times [[Differentiable function|differentiable]] at the point {{nowrap|''a'' ∈ '''R'''}}. Then there exists a function  {{nowrap|''h<sub>k</sub>'' : '''R''' → '''R'''}} such that

<math display="block"> f(x) = \sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i + h_k(x)(x-a)^k,</math>
and
<math display="block">\lim_{x\to a} h_k(x) = 0.</math>
This is called the '''[[Peano]] form of the remainder'''.
}}

The polynomial appearing in Taylor's theorem is the '''<math display="inline">\boldsymbol{k}</math>-th order Taylor polynomial'''

<math display="block">P_k(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k </math>

of the function ''f'' at the point ''a''. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function  {{nowrap|''h<sub>k</sub>'' : '''R''' → '''R'''}} and a <math display="inline">k</math>-th order polynomial ''p'' such that

<math display="block"> f(x) = p(x) + h_k(x)(x-a)^k, \quad \lim_{x\to a} h_k(x) = 0 ,</math>

then ''p''&nbsp;=&nbsp;''P<sub>k</sub>''. Taylor's theorem describes the asymptotic behavior of the '''remainder term'''

<math display="block"> R_k(x) = f(x) - P_k(x),</math>

which is the [[approximation error]] when approximating ''f'' with its Taylor polynomial. Using the [[little-o notation]], the statement in Taylor's theorem reads as

<math display="block">R_k(x) = o(|x-a|^{k}), \quad x\to a.</math>

=== Explicit formulas for the remainder ===

Under stronger regularity assumptions on ''f'' there are several precise formulas for the remainder term ''R<sub>k</sub>'' of the Taylor polynomial, the most common ones being the following.

{{math theorem
| name = Mean-value forms of the remainder
| math_statement = Let {{nowrap|''f'' : '''R''' → '''R'''}}  be ''k''&nbsp;+&nbsp;1 times [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math> with ''f''{{i sup|(''k'')}} [[continuous function|continuous]] on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>.<ref>The hypothesis of ''f''{{i sup|(''k'')}} being [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] between <math display=inline>a</math> and <math display=inline>x</math> is ''not'' redundant. Although ''f'' being ''k''&nbsp;+&nbsp;1 times [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math> does imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] on the [[open interval|''open'' interval]] between <math display=inline>a</math> and <math display=inline>x</math>, it does ''not'' imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] between <math display=inline>a</math> and <math display=inline>x</math>, i.e. it does not imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] at the ''endpoints'' of that interval. Consider, for example, the [[Function (mathematics)|function]] {{nowrap|''f'' : [0,1] → '''R'''}} defined to equal <math> \sin(1/x)</math> on <math>(0,1]</math> and with <math>f(0)=0</math>. This is not [[continuous function|continuous]] at ''0'', but is [[continuous function|continuous]] on <math>(0,1)</math>. Moreover, one can show that this [[Function (mathematics)|function]] has an [[antiderivative]]. Therefore that [[antiderivative]] is [[Differentiable function|differentiable]] on <math>(0,1)</math>, its [[derivative]] (the function ''f'') is [[continuous function|continuous]] on the [[open interval|''open'' interval]] <math>(0,1)</math>, but its [[derivative]] ''f'' is ''not'' [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] <math>[0,1]</math>. So the theorem would not apply in this case.</ref> Then

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1} </math>

for some real number <math display="inline">\xi_L</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Joseph Louis Lagrange|Lagrange]] form'''<ref>{{harvnb|Kline|1998|loc=§20.3}}; {{harvnb|Apostol|1967|loc=§7.7}}.</ref> of the remainder.

Similarly,

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_C)}{k!}(x-\xi_C)^k(x-a) </math>

for some real number <math display="inline">\xi_C</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Augustin Louis Cauchy|Cauchy]] form'''<ref>{{harvnb|Apostol|1967|loc=§7.7}}.</ref> of the remainder.

Both can be thought of as specific cases of the following result: Consider <math>p>0</math>

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_S)}{k!}(x-\xi_S)^{k+1-p}\frac{(x-a)^p}{p} </math>
for some real number <math display="inline">\xi_S</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Oskar Schlömilch|Schlömilch]] form''' of the remainder (sometimes called the '''Schlömilch-[[Édouard Roche|Roche]]'''). The choice <math display="inline">p=k+1</math> is the Lagrange form, whilst the choice <math display="inline">p=1</math> is the Cauchy form.
}}

These refinements of Taylor's theorem are usually proved using the [[mean value theorem]], whence the name. Additionally, notice that this is precisely the [[mean value theorem]] when <math display="inline">k=0</math>. Also other similar expressions can be found. For example, if ''G''(''t'') is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between <math display="inline">a</math> and <math display="inline">x</math>, then

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)} </math>

for some number <math display="inline">\xi</math> between <math display="inline">a</math> and <math display="inline">x</math>. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using [[mean value theorem#Cauchy's mean value theorem|Cauchy's mean value theorem]]. The Lagrange form is obtained by taking <math>G(t)=(x-t)^{k+1}</math> and the Cauchy form is obtained by taking <math>G(t)=t-a</math>.

{{anchor|Integral form of the remainder}}The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of [[Lebesgue integral|Lebesgue integration theory]] for the full generality. However, it holds also in the sense of [[Riemann integral]] provided the (''k''&nbsp;+&nbsp;1)th derivative of ''f'' is continuous on the closed interval [''a'',''x''].

{{math theorem|name=Integral form of the remainder<ref>{{harvnb|Apostol|1967|loc=§7.5}}.</ref> |math_statement=Let <math display=inline>f^{(k)}</math> be [[absolutely continuous]] on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>. Then

<math display="block"> R_k(x) = \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. </math>

}}

Due to the [[absolutely continuous|absolute continuity]] of ''f''{{i sup|(''k'')}} on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>, its derivative ''f''{{i sup|(''k''+1)}} exists as an ''L''{{i sup|1}}-function, and the result can be [[#Derivation for the integral form of the remainder|proven]] by a formal calculation using the [[fundamental theorem of calculus]] and [[integration by parts]].

=== Estimates for the remainder ===

It is often useful in practice to be able to estimate the remainder term appearing in the Taylor approximation, rather than having an exact formula for it.  Suppose that ''f'' is {{nowrap|(''k'' + 1)}}-times continuously differentiable in an interval ''I'' containing ''a''. Suppose that there are real constants ''q'' and ''Q'' such that

<math display="block">q\le f^{(k+1)}(x)\le Q</math>

throughout ''I''. Then the remainder term satisfies the inequality<ref>{{harvnb|Apostol|1967|loc=§7.6}}</ref>

<math display="block">q\frac{(x-a)^{k+1}}{(k+1)!}\le R_k(x)\le Q\frac{(x-a)^{k+1}}{(k+1)!},</math>

if {{nowrap|''x'' > ''a''}}, and a similar estimate if {{nowrap|''x'' < ''a''}}. This is a simple consequence of the Lagrange form of the remainder. In particular, if

<math display="block">|f^{(k+1)}(x)|\le M</math>

on an interval {{nowrap|1=''I'' = (''a'' − ''r'',''a'' + ''r'')}} with some <math>r > 0</math> , then

<math display="block">|R_k(x)|\le M\frac{|x-a|^{k+1}}{(k+1)!}\le M\frac{r^{k+1}}{(k+1)!}</math>

for all {{nowrap|''x''∈(''a'' − ''r'',''a'' + ''r'').}} The second inequality is called a [[uniform convergence|uniform estimate]], because it holds uniformly for all ''x'' on the interval {{nowrap|(''a'' − ''r'',''a'' + ''r'').}}

=== Example ===

[[File:Expanimation.gif|thumb|400px|right|Approximation of <math display="inline">e^x</math> (blue) by its Taylor polynomials <math>P_k</math> of order <math display="inline">k=1,\ldots,7</math> centered at <math display="inline">x=0</math> (red).]] Suppose that we wish to find the approximate value of the function <math display="inline">f(x)=e^x</math> on the interval <math display="inline">[-1,1]</math> while ensuring that the error in the approximation is no more than 10<sup>−5</sup>. In this example we pretend that we only know the following properties of the exponential function:

{{NumBlk|:|<math>e^0=1, \qquad \frac{d}{dx} e^x = e^x, \qquad e^x>0, \qquad x\in\R.</math>|{{EquationRef|★}}}}

From these properties it follows that <math display="inline">f^{(k)}(x)=e^x</math> for all <math display="inline">k</math>, and in particular, <math display="inline">f^{(k)}(0)=1</math>. Hence the ''<math display="inline">k</math>''-th order Taylor polynomial of <math display="inline">f</math> at <math display="inline">0</math> and its remainder term in the Lagrange form are given by

<math display="block"> P_k(x) = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^k}{k!}, \qquad R_k(x)=\frac{e^\xi}{(k+1)!}x^{k+1},</math>

where <math display="inline">\xi</math> is some number between 0 and ''x''. Since ''e''<sup>''x''</sup> is increasing by ({{EquationNote|★}}), we can simply use <math display="inline">e^x \leq 1</math> for <math display="inline">x \in [-1,0]</math> to estimate the remainder on the subinterval <math>[-1,0]</math>. To obtain an upper bound for the remainder on <math>[0,1]</math>, we use the property <math display="inline">e^\xi <e^x</math> for <math display="inline">0<\xi<x</math> to estimate

<math display="block"> e^x = 1 + x + \frac{e^\xi}{2}x^2 < 1 + x + \frac{e^x}{2}x^2, \qquad 0 < x\leq 1 </math>

using the second order Taylor expansion. Then we solve for ''e<sup>x</sup>'' to deduce that

<math display="block"> e^x \leq \frac{1+x}{1-\frac{x^2}{2}} = 2\frac{1+x}{2-x^2} \leq 4, \qquad 0 \leq x\leq 1 </math>

simply by maximizing the [[numerator]] and minimizing the [[denominator]]. Combining these estimates for ''e<sup>x</sup>'' we see that

<math display="block"> |R_k(x)| \leq \frac{4|x|^{k+1}}{(k+1)!} \leq \frac{4}{(k+1)!}, \qquad -1\leq x \leq 1, </math>

so the required precision is certainly reached, when

<math display="block"> \frac{4}{(k+1)!} < 10^{-5} \quad \Longleftrightarrow \quad 4\cdot 10^5 < (k+1)!  \quad \Longleftrightarrow \quad k \geq 9.
</math>

(See [[factorial]] or compute by hand the values <math display="inline">9! =362880</math> and <math display="inline">10! =3628800</math>.) As a conclusion, Taylor's theorem leads to the approximation

<math display="block"> e^x = 1+x+\frac{x^2}{2!} + \cdots + \frac{x^9}{9!} + R_9(x), \qquad |R_9(x)| < 10^{-5}, \qquad -1\leq x \leq 1. </math>

For instance, this approximation provides a [[decimal representation|decimal expression]] <math>e \approx 2.71828</math>, correct up to five decimal places.