91Ó°ÊÓ

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

The definition of the Maclaurin series is defined.

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

$(1-x{)}^{-1}=\underset{n=0}{\overset{∞}{∑}}\left(\begin{array}{c}-1\\ n\end{array}\right)(-1{)}^n{x}^n$

$=1+x+x^2+x^3+x^4+x^5+…$

State the Maclaurin series.

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

$(1-x{)}^{-1}=\underset{n=0}{\overset{∞}{∑}}\left(\begin{array}{c}-1\\ n\end{array}\right)(-1{)}^n{x}^n$

$=1+x+x^2+x^3+x^4+x^5+…$

Use the first Maclaurin series and evaluate.

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

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

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

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

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

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

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

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