Maclaurin series for $\frac{\sin{x}}{1-2x}$

Question

The function is $$ \frac{\sin{x}}{1-2x},$$

we have to find its Maclaurin series using composite functions

I know the other method of manually calculating the derivatives,but the question specifically asks for composite functions. I tried taking the Maclaurin series of sin x and then divide by 1 - 2x, but that didn't give me the right answer so how do I use composite functions to calculate this?

Answer 1

You need to expand $(1-2x)^{-1}$ as well. (I am not sure what you mean by using composite functions but this should suffice.

$\sin(x)=x-x^3/3!+x^5/5!....=x-x^3/6+x^5/120-...$

and

$(1-2x)^{-1}=1+(-1)(-2x)+\frac{(-1)(-2)}{2!}(-2x)^2...=1+2x+4x^2+...$

Multiply these out to get

$\frac{\sin(x)}{1-2x}=x+2x^2+4x^3-\frac{1}{6}x^3+...=x+2x^2+\frac{23}{6} x^3+...$

And so on...

Answer 2

Another method consists performing the division by increasing power order of the expansion of $\sin x$ by $1-2x$, up to the required order of expansion:

\begin{array}{rclrrc} &&x+2x^2&{}+\frac{23}6x^3&{}+\frac{23}3x^4&{}+\cdots\\ \llap{1-2x} &\!\!\llap{\Big)}\!\!&x&{}-\frac16x^3&+\frac1{120}x^5&{}+\cdots\\ &&\llap-x+2x^2&\\ \hline &&2x^2{}&{}-\frac16x^3&{}+\frac1{120}x^5&{}+\cdots&{}+\cdots\\ &&-2x^2&{}+4x^3 \\ \hline &&&\frac{23}6x^3&{}+\frac1{120}x^5&{}+\cdots \\ &&&{}-\frac{23}6x^3&{}+\frac{23}{3}x^4 \end{array}