Question

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

$\frac{1+x}{1-x}$

Short answer

Ans: The Maclaurin series of the function $f\left(x\right)=1+2\sum _{k=1}^{\infty }\frac{2·3·4\cdots k}{k!}{x}^k$

Step-by-step solution

Step 1. Given information:

$f\left(x\right)=\frac{1+x}{1-x}$

Step 2. Finding the general form  Maclaurin series:  

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

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

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

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function $f\left(x\right)=\frac{1+x}{1-x}$

$\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& \frac{1+x}{1-x}& 1& 1\\ 1& \frac{2}{(1-x{)}^2}& 2& 2\\ 2& \frac{2·2}{(1-x{)}^3}& 2·2& \frac{2·2}{2!}\\ 3& \frac{2·2·3}{(1-x{)}^4}& 2·2·3& \frac{2·2·3}{3!}\\ 4& \frac{2·2·3·4}{(1-x{)}^5}& 2·2·3·4& \frac{2·2·3·4}{4!}\\ k& \frac{2[2·3·4\cdots k]}{(1-x{)}^{k+1}}& 2[2·3·4\cdots k]& \frac{2[2·3·4\cdots k]}{k!}\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=\frac{1+x}{1-x}$is $1+2·x+\frac{2·2}{2!}{x}^2+\frac{2·2·3}{3!}{x}^3+\frac{2·2·3·4}{4!}{x}^4+\dots +\frac{2[2·3·4\cdots k]}{k!}{x}^k$

Or, we can write as

$f\left(x\right)=1+2\sum _{k=1}^{\infty }\frac{2·3·4\cdots k}{k!}{x}^k$