Taylor series

Differentiation, derivatives and Taylor approximations: Taylor approximations

Taylor series

The derivative of the exponential function $e^x$ is the function itself. We take $a=0, x>0$ and find with Taylor's theorem

$e^x=1+x+\frac{1}{2!}\!x^2+\frac{1}{3!}\!x^3+\cdots + \frac{1}{k!}\!x^k+R_k(x)$ where $R_k(x)=\frac{1}{(k+1)!}e^{\xi} x^{k+1}$ for some $0<\xi<x$ .

We now take $x=1, k=6$ and take into account that the base of the natural logarithm is less than 3 (ie $e<3$ ). So then we get

$e=1+ 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+R_6(1)$ with $0<R_6(1)<\frac{3}{5040}=\frac{1}{1680}<0.0006$ . So we have approximated $e$ with $\frac{1957}{720}\approx 2.7180\ldots$ up to four significant digits.

In this last example, you could of course take as many terms as you want and you could write the following series expansion:

$\begin{aligned}e^x&=1+x+\frac{1}{2!}\!x^2+\frac{1}{3!}\!x^3+ \frac{1}{4!}\!x^4+\frac{1}{5!}\!x^5+\cdots\\ \\ &=\sum_{k=0}^{\infty}\frac{1}{k!}x^k\end{aligned}$ This is called the Taylor series of $e^x$ about $x=0$ .

Note that the following is true in particular:

$e= \sum_{k=0}^{\infty}\frac{1}{k!}$

The Taylor series of sine and cosine about the origin are:

$\begin{aligned}\sin x&=x-\frac{1}{3!}\!x^3+ \frac{1}{5!}\!x^5-\frac{1}{7!}\!x^7+\cdots\\ \\ &=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ \\ \cos x &=1-\frac{1}{2!}\!x^2+ \frac{1}{4!}\!x^4-\frac{1}{6!}\!x^6+\cdots\\ \\ &=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}x^{2k} \end{aligned}$

The Taylor series for the exponential function, sine and cosine function are not only valid for all values near $x=0$ , but even for all values of $x$ . This is not always the case.

For the Taylor series of $\ln(1+x)$ we have

$\begin{aligned}\ln(1+x) &= x -\frac{1}{2}\!x^2+\frac{1}{3}\!x^3 - \frac{1}{4}\!x^4 +\cdots\\ \\ &= \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} x^k\end{aligned}$ but this is only valid for $-1<x\le 1$ .

You can numerically approximate $\ln(2)$ by adding the first terms of $1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4}+\cdots$ But a really good approximation is only obtained by using a lot of terms: even after using 1000 terms you only get two correct decimals.

It can be done in a more clever way by taking $x=-\tfrac{1}{2}$ and realizing that

$\ln\bigl(1+(-\tfrac{1}{2})\bigr)=\ln(\tfrac{1}{2})=-\ln(2)\tiny.$ Then you get after 10 terms in the Taylor series an approximation of $\ln(2)$ that is correct in four decimals.