Taylor Series of $\sin x/(1-x)$ Ιs there any fast way to calculate the first four non-zero terms a Taylor Series $\dfrac {\sin x}{1-x}$ at $x=0$ without making big derivatives calculations?
I know that $$\sin x = x- \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} \dots$$
and $$\frac{1}{1-x} = 1 + x + x^2 + x^3 \dots$$
Can we combine them together?
 A: Maybe
$$(1-x)(a_0+a_1x+a_2x^2+a_3x^3\cdots)=x-\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}\cdots$$
$$a_0=0\\a_1-a_0=1\\a_2-a_1=0\\a_3-a_2=-\frac1{3!}\\a_4-a_3=0\\a_5-a_4=\frac1{5!}\\\cdots$$
so that
$$a_{2k+2}=a_{2k+1}=\sum_{i=0}^k\frac{(-1)^i}{(2i+1)!}.$$
The coefficients are the successive Taylor approximations of $\sin(1)$.
A: It is possible to multiply series together. It is a little bit tedious, but you are basically just going to use the distributive property. However, you can't just distribute normally, otherwise you'll have an infinite collection of infinite sums - we want to distribute in a way which lets us read off the terms of the series.
To do this, we will basically compute the terms order by order. That is to say, given the two series, first we will multiply out all the terms which give order 0 exponents (just the first two), then we will multiply out all the terms which multiply to order 1 (the constant of the first times the linear of the second and vice versa), then the terms which multiply to order two, and so on, and so forth.
I'm not the best with LaTeX, and I don't think it'll be clear if I just show how to do it. Instead, I've attached a link here which works examples.
http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx
(scroll down to about half way, then the explanation begins. Alternatively, just ctrl+f "multiplication")
In any case, wolfram computes the series just fine as well.
http://www.wolframalpha.com/input/?i=taylor+series+sin+%28x%29%2F%281-x%29
A: If you don't want to multiply the two series together and you want to avoid heavy differentiation (and if you actually needed more terms in the series) you could try the following:
$$y(1-x)=\sin x$$
$$y'(1-x)-y=\cos x$$
$$y''(1-x)-2y'=-\sin x$$
$$y'''(1-x)-3y''=-\cos x$$
$$y^{(4)}(1-x)-4y'''=\sin x$$
And so on...
