Chapter Seven – Taylor Series Remainder Test

A formal way to test the accuracy of a Taylor polynomial approximation is to employ the Taylor Remainder test. By adding a remainder term to our Taylor polynomial approximation, we in effect convert it into an equation,

Our function [latex]f(x) = p(x) + remainder term[/latex].  Written more compactly we have  [latex]f_n(x) = p_n(x) +  r_n(x)[/latex] . This remainder term becomes the difference between [latex]f(x)[/latex] at a particular point and [latex]p(x)[/latex] at that same value of x.

In the above example we ran our polynomial out to the ninth-degree term.

[latex]r_n(x)[/latex] actually looks like the next higher degree term:

[latex]\large r_n(x) = \frac {f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}[/latex] where c is between a and x

The question we ask is what value for c should we use. The answer in this case is to solve the remainder twice for the endpoints of the range we are interested in. In this case we want to know how accurate c will be between [latex]0[/latex] and [latex]\frac {\pi}{10}[/latex].

This will provide a range of possible values between [latex]0[/latex] and [latex]\large \frac {\pi}{10}[/latex]

[latex]\large f^{(10)}(c) = -sin(c)[/latex]

For [latex]\large c = 0[/latex] [latex]\large r_9(0) = \frac {-sin(0)}{(10)!} (0 - 0)^{10} = 0[/latex]

For [latex]\large c = \frac {\pi}{10}[/latex]  [latex]r_9(\frac {\pi}{10}) =\frac {-sin(\frac{\pi}{10})}{(10)!} (\frac {\pi}{10} - 0)^{10} = -0.000000000000797[/latex] we drop the negative as it’s a matter of distance, not direction.

This bounds the possible error of our approximation: [latex]0 \le r_{10} \le 0.000000000000797[/latex] [latex](7.97 * 10^{-13})[/latex]