Chapter 1: Q17MP (page 45)

Find the Maclaurin series of the following functions.
$e^{- 1 \sqrt{1 - x^{2}}}$

Short Answer

Expert verified

The Maclaurin series of $e^{- 1 \sqrt{1 - x^{2}}}$ is $1 + \frac{x^{2}}{2} + \frac{x^{4}}{4} + \frac{7 x^{6}}{48} + \dots$ .

Step by step solution

Maclaurin series and the given function:

Function is $e^{- 1 \sqrt{1 - x^{2}}}$ ,

The Maclaurin series of $e^{x}$ is expressed as follows:

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$

The Maclaurin series of is expressed as follows:

role="math" localid="1658895933390" ${(1 + x)}^{1 / 2} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + \dots$

Calculation by the Maclaurin series:

Replace x by $- x^{2}$ in the Maclaurin series of ${(1 + x)}^{1 / 2}$ as follows:

${(1 - x^{2})}^{\frac{1}{2}} = 1 - \frac{1}{2} x^{2} - \frac{1}{8} x^{4} - \frac{1}{16} x^{6} + \dots {(1 - x^{2})}^{\frac{1}{2}} = \frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots$

Therefore, the Maclaurin series of $1 - {(1 - x^{2})}^{\frac{1}{2}}$ is $\frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots$

Substitute $\frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots$ for x in expansion of $e^{x}$ as follows:

role="math" localid="1658897078635" $e^{- 1 \sqrt{1 - x^{2}}} = [\begin{matrix} 1 + [\frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots] + \frac{{[\frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots]}^{2}}{2!} \\ + \frac{[\frac{1}{2} x^{2} + \frac{1}{8} x^{4} + \frac{1}{16} x^{6} + \dots]}{3!} \end{matrix}] e^{- 1 \sqrt{1 - x^{2}}} = [1 + \frac{192 x^{2} + 96 x^{4} + 56 x^{6}}{384} + . . . .] = 1 + \frac{x^{2}}{2} + \frac{x^{4}}{4} + \frac{7 x^{6}}{48} + . . . . . .$

Hence, the Maclaurin series of $e^{- 1 \sqrt{1 - x^{2}}}$ is $1 + \frac{x^{2}}{2} + \frac{x^{4}}{4} + \frac{7 x^{6}}{48} + . . . . . .$

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Find the Maclaurin series of the following functions.
$e^{- 1 \sqrt{1 - x^{2}}}$

Short Answer

Step by step solution

Maclaurin series and the given function:

Calculation by the Maclaurin series:

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