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Q 62

Expert-verifiedFound in: Page 680

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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

$(1+x{)}^{2/3}$

The maclaurin series for the given function is

$(1+x{)}^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots $

We have been given

$(1+x{)}^{2/3}$

to find the maclaurin series by using binomial 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 $\sum _{k=0}^{\mathrm{\infty}}\u200a\left(\begin{array}{l}p\\ k\end{array}\right){x}^{k}$

where the binomial coefficient is

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

So for the given function $f\left(x\right)=(1+x{)}^{\frac{2}{3}}$ binomial series is ,

$(1+x{)}^{\frac{2}{3}}=\sum _{k=0}^{\mathrm{\infty}}\u200a\left(\begin{array}{l}\frac{2}{3}\\ k\end{array}\right){x}^{k}$

implies that ,

$\begin{array}{r}(1+x{)}^{\frac{2}{3}}=\left(\begin{array}{c}\frac{2}{3}\\ 0\end{array}\right){x}^{0}+\left(\begin{array}{c}\frac{2}{3}\\ 1\end{array}\right){x}^{1}+\left(\begin{array}{c}\frac{2}{3}\\ 2\end{array}\right){x}^{2}+\left(\begin{array}{c}\frac{2}{3}\\ 3\end{array}\right){x}^{3}+\cdots \\ =1+\frac{2}{3}x+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2!}{x}^{2}+\frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{3!}{x}^{3}+\cdots \\ =1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots \end{array}$

The maclaurin series for the given function is $(1+x{)}^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\cdots $

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