Editing Taylor's theorem (section)

=== 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.