91Ó°ÊÓ

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

Since for any function $f$with derivatives of all orders at the point $x_0=0$, then the Maclurin series is

$f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(0\right)}{4!}{x}^4+…$

Or, we can write the general form Maclurin series of the function $f$is

$f\left(x\right)=\underset{n=0}{\overset{∞}{∑}}\frac{f^n\left(0\right)}{n!}{x}^n$

So, let us first construct the table of the Maclaurin series for the function $f\left(x\right)=\cos x$

$\begin{array}{|cccc|}\hline n& f^{\text{'}n}\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& \ln (1+x)& 0& 0\\ 1& \frac{1}{1+x}& 1& 1\\ 2& -\frac{1}{(1+x{)}^2}& -1& -\frac{1}{2!}\\ 3& \frac{2}{(1+x{)}^3}& 2& \frac{2}{3!}\\ 4& -\frac{6}{(1+x{)}^4}& -6& -\frac{6}{4!}\\ k& \frac{(-1{)}^{k-1}(k-1)!}{(1+x{)}^k}& (-1{)}^{k-1}(k-1)!& \frac{(-1{)}^{k-1}(k-1)!}{k!}\\ \hline\end{array}$

Therefore, the Maclaurin series for the function $f\left(x\right)=\ln (1+x)$is

$0+1·x+\frac{-1}{2!}{x}^2+\frac{2}{3!}{x}^3+\frac{(-6)}{4!}{x}^4+…$

Or, we can write it as

$f\left(x\right)=\ln 1+\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^{k-1}(k-1)!}{k!}{x}^k$

Since $\ln 1=0$, so the Maclaurin series for the function $f\left(x\right)=\ln (1+x)$can also be written as $f\left(x\right)=\underset{k=1}{\overset{∞}{∑}}\frac{(-1{)}^{k-1}(k-1)!}{k!}{x}^k$