Question

Find the Maclaurin series of the following functions.

${\mathbf{e}}^{\mathbf{-}\mathbf{1}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}$

Short answer

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

Step-by-step solution

Maclaurin series and the given function:

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

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

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

The Maclaurin series of is expressed as follows:

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

Calculation by the Maclaurin series:

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

$$\begin{gathered} {\left(1-{\mathrm{x}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1-\frac{1}{2}{\mathrm{x}}^2-\frac{1}{8}{\mathrm{x}}^4-\frac{1}{16}{\mathrm{x}}^6+\cdots \\ {(1-{\mathrm{x}}^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{2}{\mathrm{x}}^2+\frac{1}{8}{\mathrm{x}}^4+\frac{1}{16}{\mathrm{x}}^6+\cdots \end{gathered}$$

Therefore, the Maclaurin series of $1-{\left(1-{\mathrm{x}}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$is$\frac{1}{2}{\mathrm{x}}^2+\frac{1}{8}{\mathrm{x}}^4+\frac{1}{16}{\mathrm{x}}^6+\cdots$

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

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

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