example of use of Taylor’s theorem

In this entry we use Taylor’s Theorem in the following form:

Theorem 1 (Taylor’s Theorem: Bounding the Error).

Suppose $f$ and all its derivatives are continuous. If $T_{n}(x)$ is the $n$ -th Taylor polynomial of $f(x)$ around $x=a$ , then the error, or the difference between the real value of $f$ and the values of $T_{n}(x)$ is given by:

|E_{n}(x)|=|f(x)-T_{n}(x)|\\leq\\frac{M}{(n+1)!}|x-a|^{n+1}

where $M$ is the maximum value of $f^{(n+1)}$ (the $n+1$ -th derivative of $f$ ) in the interval between $a$ and $x$ .

Example 2.

Suppose we want to approximate $e$ using the Taylor polynomial of degree 4, $T_{4}(x)$ , around $x=0$ for the function $e^{x}$ . We know that

T_{4}(x)=1+x+x^{2}/2+x^{3}/3!+x^{4}/4!

so we are asking how close are $e$ and $T_{4}(1)=1+1+1/2+1/6+1/24$ . In order to use the formula in the theorem, we just need to find $M$ , the maximum value of the $4$ th derivative of $e^{x}$ between $a=0$ and $x=1$ . Since $f^{(4)}=e^{x}$ and $e^{x}$ is strictly increasing, the maximum in $(0,1)$ happens at $x=1$ . Thus $M=e$ which is a number, say, less than $3$ . Therefore:

|E_{4}(1)|=|f(1)-T_{4}(1)|=|e-(1+1+1/2+1/6+1/24)|\\leq\\frac{M}{5!}|1-0|^{5}=%\\frac{M}{5!}<\\frac{3}{5!}=0.025

Thus the approximation has an error of less than $0.025$ . Actually, if we use a calculator we obtain that the error is exactly $0.0099$ . But, of course, the whole point of the theorem is not to use a calculator.

Example 3.

What Taylor polynomial $T_{n}(x)$ (what $n$ ) should we use to approximate $e$ within $0.0001$ ? As above, we will be using the Taylor polynomial $T_{n}(x)$ for $e^{x}$ , evaluated at $x=1$ . Thus, we want the error $|E_{n}(1)|<0.0001$ . Notice all derivatives are $e^{x}$ and the maximum happens at $x=1$ , where $e^{1}=e$ , so for all derivatives $M<3$ . Hence by the theorem:

|E_{n}(1)|=|f(1)-T_{n}(1)|=|e-(1+1+1/2+\\ldots+1/n!)|\\leq\\frac{M}{(n+1)!}|1-0|^%{n+1}=\\frac{M}{(n+1)!}<\\frac{3}{(n+1)!}

So we need $3/(n+1)!<0.0001$ . Try several values of $n$ until that is satisfied:

3/2=1.5,\\ 3/3!=0.5,\\ 3/4!=0.125,\\ 3/5!=0.025,\\ 3/6!=0.00416

3/7!=0.00059,\\quad 3/8!=0.00007

Thus $n=7$ should work. So we just need $T_{7}(x)$ , or add $1+1+1/2+\\ldots+1/7!$ .