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Q 62

Expert-verified

Calculus

Found in: Page 680

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

We have been given

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

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 $g (x) = (1 + x)^{p}$

is called the binomial series which is given by $\sum_{k = 0}^{\infty} (\begin{array}{l} p \\ k \end{array}) x^{k}$

where the binomial coefficient is

$(\begin{array}{l} p \\ k \end{array}) = \{\begin{array}{rr} \frac{p (p - 1) (p - 2) \dots (p - k + 1)}{k!}, if k > 0 \\ 1, if k = 0 \end{array}$

Step 3. Binomial series for the given function is

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

$(1 + x)^{\frac{2}{3}} = \sum_{k = 0}^{\infty} (\begin{array}{l} \frac{2}{3} \\ k \end{array}) x^{k}$

implies that ,

$\begin{aligned} (1 + x)^{\frac{2}{3}} = (\begin{matrix} \frac{2}{3} \\ 0 \end{matrix}) x^{0} + (\begin{matrix} \frac{2}{3} \\ 1 \end{matrix}) x^{1} + (\begin{matrix} \frac{2}{3} \\ 2 \end{matrix}) x^{2} + (\begin{matrix} \frac{2}{3} \\ 3 \end{matrix}) x^{3} + \dots \\ = 1 + \frac{2}{3} x + \frac{\frac{2}{3} (- \frac{1}{3})}{2!} x^{2} + \frac{\frac{2}{3} (- \frac{1}{3}) (- \frac{4}{3})}{3!} x^{3} + \dots \\ = 1 + \frac{2}{3} x - \frac{1}{9} x^{2} + \frac{4}{81} x^{3} + \dots \end{aligned}$

Step 4. The maclaurin series for given function is

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

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