Question

$\frac{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{x}}$

Short answer

$\frac{1-\sin \left(x\right)}{1-x}=1+\frac{x^3}{6}+\frac{x^4}{6}+\frac{19x^5}{120}+\dots$

Step-by-step solution

Given information

The Maclaurin expansion of the function $\frac{1-\sin x}{1-x}$is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$sin\left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$(1-x{)}^{-1}=\sum _{n=0}^{\infty }\left(\begin{array}{c}-1\\ n\end{array}\right)(-1{)}^n{x}^n$

$=1+x+x^2+x^3+x^4+x^5+\dots$

Paer-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$\sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$(1-x{)}^{-1}=\sum _{n=0}^{\infty }\left(\begin{array}{c}-1\\ n\end{array}\right)(-1{)}^n{x}^n$

$=1+x+x^2+x^3+x^4+x^5+\dots$

Part-B Step 3: Use the Maclaurin series to evaluate

Use the first Maclaurin series and evaluate.

$1-\sin \left(x\right)=1-\left[x-\frac{x^3}{6}+\frac{x^5}{120}-\dots \right]$

$1-\sin \left(x\right)=1-x+\frac{x^3}{6}-\frac{x^5}{120}+\dots$

Part-C Step 3: Evaluate the denominator and combine to find the answer

Use the second Maclaurin series to find the expansion and evaluate the find the answer.

$\frac{1-\sin \left(x\right)}{1-x}=(1-\sin (x\left)\right)(1-x{)}^{-1}$

$=\left[1-x+\frac{x^3}{6}-\frac{x^5}{120}+\dots \right]\times \left[1+x+x^2+x^3+x^4+x^5+\dots \right]$

$=\left[\begin{array}{l}\left[1+x+x^2+x^3+x^4+x^5+\dots \right]-x·\left[1+x+x^2+x^3+x^4+x^5+\dots \right]+\\ \frac{x^3}{6}·\left[1+x+x^2+x^3+x^4+x^5+\dots \right]-\frac{x^5}{120}·\left[1+x+x^2+x^3+x^4+x^5+\dots \right]\end{array}\right]$

$\frac{1-\sin \left(x\right)}{1-x}=1+\frac{x^3}{6}+\frac{x^4}{6}+\frac{19x^5}{120}+\dots$

Thus, the final answer is$\frac{1-\sin \left(x\right)}{1-x}=1+\frac{x^3}{6}+\frac{x^4}{6}+\frac{19x^5}{120}+\dots$