Question

Use power series to evaluate the function at the given point. Compare with computer results, using the computer to find the series, and also to do the problem without series. Resolve any disagreement in results (see Example 1).${\mathbf{e}}^{\mathbf{arcsin}\mathbf{}\mathbf{x}}\mathbf{+}\mathbf{In}\mathbf{\left(}\frac{\mathbf{1}\mathbf{-}\mathbf{x}}{\mathbf{e}}\mathbf{\right)}\mathbf{at}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0003}$.

Short answer

The value of the function is$-3.376\times {10}^{-16}$ .

Step-by-step solution

Given Information

The function is ${\mathrm{e}}^{\mathrm{arc}\sin \mathrm{x}}+\mathrm{In}\left(\frac{1-\mathrm{x}}{\mathrm{e}}\right)$

Definition of the Maclaurin Series

A Maclaurin series is a function with an expansion series that gives the sum of the function's derivatives.

Find the value

The formula is mentioned below as:

$$\begin{gathered} {\sin }^{-1}\left(x\right)=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+... \\ {e}^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \end{gathered}$$

The function becomes as follows:

$e^{arc\sin \left(x\right)}=1+\left[x+\frac{x^3}{6}+\frac{3x^5}{40}+...\right]\frac{{\left[x+{\displaystyle \frac{x^3}{6}}+{\displaystyle \frac{3x^5}{40}}+...\right]}^2}{2!}+\frac{{\left[x+\frac{x^3}{6}+\frac{3x^5}{40}+...\right]}^2}{3!}+...$

The value of the function for x=0.0003, and we get,

$$\begin{gathered} e^{arc\sin \left(0.0003\right)}=1+\left[0.0003+\frac{0.{0003}^3}{6}+\frac{3\times 0.{0003}^5}{40}+...\right]\frac{{\left[0.0003+\frac{0.{0003}^3}{6}+\frac{3\times 0.{0003}^5}{40}+...\right]}^2}{2!} \\ \frac{{\left[0.0003+\frac{0.{0003}^3}{6}+\frac{3\times 0.{0003}^5}{40}+...\right]}^2}{3!}+ \end{gathered}$$

localid="1658547873588" $$\begin{gathered} \mathrm{for}\mathrm{In}\left(\frac{1-x}{e}\right),\mathrm{The}\mathrm{series}\mathrm{is}\mathrm{given}\mathrm{below}\mathrm{as}: \\ In\left(\frac{1-x}{e}\right)=In\left(1-x\right)-In\left(e\right) \\ =In\left(1-x\right)-1 \\ \mathrm{In}\left(1-x\right)=-\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}+... \\ \mathrm{In}{\left(\frac{1-\mathrm{x}}{\mathrm{e}}\right)}_{\mathrm{x}=0.0003}=-1-0.0003-\frac{0.{0003}^2}{2}-\frac{0.{0003}^3}{3}+... \end{gathered}$$

The sum of the function is given below as:

localid="1658548346428" $$\begin{gathered} e^{arc\sin \left(0.0003\right)}+In\left(\frac{1-0.0003}{e}\right)=1+\left[0.0003+\frac{0.{0003}^3}{6}+\frac{3\times 0.{0003}^5}{40}+...\right] \\ +\frac{{\left[0.0003+{\displaystyle \frac{0.{0003}^3}{6}}+{\displaystyle \frac{3\times 0.{0003}^5}{40}}+...\right]}^2}{2!} \\ +\frac{{\left[0.0003+{\displaystyle \frac{0.{0003}^3}{6}}+{\displaystyle \frac{3\times 0.{0003}^5}{40}}+...\right]}^3}{3!} \\ +...+\left(-1-0.0003-\frac{0.{0003}^2}{2}-\frac{0.{0003}^3}{3}+...\right) \\ {e}^{arc\sin \left(0.0003\right)}+In\left(\frac{1-0.0003}{e}\right)=-3.376\times {10}^{-16} \end{gathered}$$

The value of the function is$-3.376\times {10}^{-16}$.