Question

In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .

$\sqrt[3]{1+x}$

Short answer

The maclaurin series for the given function is

$\sqrt[3]{1+x}=1+\frac{1}{3}x-\frac{1}{9}{x}^2+\frac{5}{81}{x}^3+\cdots$

Step-by-step solution

Step 1.Given information 

We have been given

$f\left(x\right)=\sqrt[3]{1+x}$

to find the maclaurin series by using binomial series

Step 2.Defining the series 

For any non- zero constant p, the Maclaurin series for the function $g\left(x\right)=(1+x{)}^p$is called the binomial series which is given by $\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}p\\ k\end{array}\right)x^k$where the binomial coefficient is $\left(\begin{array}{l}p\\ k\end{array}\right)$$\left(\begin{array}{l}p\\ k\end{array}\right)=\left\{\begin{array}{r}\frac{p(p-1)(p-2)\cdots (p-k+1)}{k!},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}k>0\\ 1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}k=0\end{array}\right.$

Step 3. Binomial series for the given function is 

so for the given function $f\left(x\right)=\sqrt[3]{1+x}$ , the binomial series is$\sqrt[3]{1+x}=\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}\frac{1}{3}\\ k\end{array}\right)x^k$

this implies that ,

$\begin{array}{r}\sqrt[3]{1+x}=\left(\begin{array}{c}\frac{1}{3}\\ 0\end{array}\right)x^0+\left(\begin{array}{c}\frac{1}{3}\\ 1\end{array}\right)x^1+\left(\begin{array}{c}\frac{1}{3}\\ 2\end{array}\right)x^2+\left(\begin{array}{c}\frac{1}{3}\\ 3\end{array}\right)x^3+\cdots \\ =1+\frac{1}{3}x+\frac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2!}{x}^2+\frac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}{x}^3+\cdots \\ =1+\frac{1}{3}x-\frac{1}{9}{x}^2+\frac{5}{81}{x}^3+\cdots \end{array}$

Step 4. Finding the maclaurin series of given function 

Hence the maclaurin series of given function is

$\sqrt[3]{1+x}=1+\frac{1}{3}x-\frac{1}{9}{x}^2+\frac{5}{81}{x}^3+\cdots$