Q. 61 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The maclaurin series for the given function is

$(1+x{)}^{-\frac{1}{2}}=1-\frac{1}{2}x+\frac{3}{8}{x}^2-\frac{5}{16}{x}^3+\cdots$

Step 1.Given information

We have been given 

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

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

$\left(\begin{array}{l}p\\ k\end{array}\right)=\left\{\begin{array}{r}\frac{p(p-1)(p-2)\cdots (p-k+1)}{k!},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }k>0\\ 1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }k=0\end{array}\right.$

Step 3. Binomial series for the given function is

So for the given function $f\left(x\right)=\frac{1}{\sqrt{1+x}}$ binomial series is ,

$(1+x{)}^{-\frac{1}{2}}=\sum _{k=0}^{\mathrm{\infty }} \left(\begin{array}{l}\left.-\frac{1}{2}\right)x^k\\ k\end{array}\right)$

implies that ,

$\begin{array}{r}(1+x{)}^{-\frac{1}{2}}=\left(\begin{array}{c}-\frac{1}{2}\\ 0\end{array}\right)x^0+\left(\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right)x^1+\left(-\frac{1}{2}\right)x^2+\left(\begin{array}{c}\left.-\frac{1}{2}\right)x^3+\cdots \\ 2\end{array}\right)\\ =1-\frac{1}{2}x+\frac{-\frac{1}{2}\left(-\frac{3}{2}\right)}{2!}{x}^2+\frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}{x}^3+\cdots \\ =1-\frac{1}{2}x+\frac{3}{8}{x}^2-\frac{5}{16}{x}^3+\cdots \end{array}$

Step 4. The maclaurin series for given function is

The maclaurin series for the given function is  $(1+x{)}^{-\frac{1}{2}}=1-\frac{1}{2}x+\frac{3}{8}{x}^2-\frac{5}{16}{x}^3+\cdots$