Compute the approximate value of In(2) using the Taylor polynomials calculated in (a). How large is the approximation error?

Consider the function,

$f\left(x\right)=\ln (x+1) \cdots \cdots \left(1\right)$

At $x=0,$ equation (1) becomes,

$\begin{array}{rcl}f\left(0\right)& =& \ln (0+1)\\ & =& \ln 1\\ & =& 0\end{array}$

Taking the derivatives of the function $f\left(x\right)$ and using $x=0,$

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{\left(x+1\right)}\\ f\text{'}\left(0\right)& =& 1\\ f\text{'}\text{'}\left(x\right)& =& \frac{-1}{{\left(x+1\right)}^2}\\ f\text{'}\text{'}\left(0\right)& =& -1\\ f\text{'}\text{'}\text{'}\left(x\right)& =& \frac{2}{{\left(x+1\right)}^3}\\ f\text{'}\text{'}\text{'}\left(0\right)& =& 2\\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)& =& \frac{-6}{{\left(x+1\right)}^4}\\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)& =& -6 \cdots \end{array}$

The Taylor polynomial of $f\left(x\right)$ of degree $n$ is given by,

$\begin{array}{rcl}\ln (x+1)& =& f\left(0\right)+\frac{f\text{'}\left(0\right)\left(x-0\right)}{1!}+\frac{f\text{'}\text{'}\left(0\right){\left(x-0\right)}^2}{2!}+\frac{f\text{'}\text{'}\text{'}\left(0\right){\left(x-0\right)}^3}{3!}+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right){\left(x-0\right)}^4}{4!}+\cdots +\frac{f^{\left(n\right)}\left(0\right){\left(x-0\right)}^n}{n!}+\cdots \\ & =& 0+\left(x\times 1\right)+\frac{x^2}{2}\left(-1\right)+\frac{x^3}{6}\left(2\right)+\frac{x^4}{24}\left(-6\right)+\cdots +\frac{f^{\left(n\right)}\left(0\right)x^n}{n!}+\cdots \\ & =& x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots +\frac{f^{\left(n\right)}\left(0\right)x^n}{n!}+\cdots \end{array}$

Then, the Taylor polynomial of $f\left(x\right)$ of degree 4 is,

$\ln (x+1)=\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{x^2}{2}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \frac{x^3}{3}\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{x^4}{4}\end{array} \cdots \cdots \left(2\right)$

Put $x=1$ in equation (2),

$\begin{array}{rcl}\ln (1+1)& =& 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\\ \ln \left(2\right)& =& 0.58333334 \cdots \cdots \left(3\right)\end{array}$