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?
Best Answer
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...