What is the Taylor expansion of x/sin(ax)?

[RedDwarf]

Hey everyone
1. Homework Statement
I want to compute the Taylor expansion (the first four terms) of $$f(x) =x/sin(ax)$$ around $$x_0 = 0$$. I am working in the space of complex numbers here.

Homework Equations

function: $$f(x) = \frac{x}{\sin (ax)}$$
Taylor expansion: $$ f(x) = \sum _{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

The Attempt at a Solution

I thought I could use the series form of sine:
$$sin(ax) = \sum (-1)^n \frac{(ax)^{2n+1}}{(2n+1)!} $$
$$x/sin(ax) = \sum (-1)^n \frac{ (2n+1)! } { a^{2n+1} }x^{-2n}$$
While this is in fact a series, this doesn't look like a Taylor expansion at all. Is there a clever way of seing the Taylor expansion without actually calculating all the derivatives by hand?
Wolfram Alpha gives a rather neat result, but I have no clue how one gets there.

 

[Orodruin]

That is not how division works. You have essentially assumed that 2=(1+3)/2 = 2/1 + 2/3 = 8/3.

I suggest you do the following:
- What is the expansion of sin(x)/x?
- What is the expansion of 1/(1+y)?
- Let y = sin(x)/x - 1.
- Remember that you only need to keep terms up to $x^3$ as you are only interested in the four first terms.

 

[RedDwarf]

That is not how division works. You have essentially assumed that 2=(1+3)/2 = 2/1 + 2/3 = 8/3.

I suggest you do the following:
- What is the expansion of sin(x)/x?
- What is the expansion of 1/(1+y)?
- Let y = sin(x)/x - 1.
- Remember that you only need to keep terms up to $x^3$ as you are only interested in the four first terms.

The expansion of $\sin (x)/x = 1-x^2/3!+x^4/5!-... = \sum (-1)^n \frac{x^{2n}}{(2n+1)!}$
The expansion of $1/(1+y) = 1-y+y^2-y^3+... = \sum (-1)^n y^n$
Then, if I set $y = \sin (x)/x -1$:
$x/sin(x) = 1- (sin(x)/x-1)+(sin(x)/x-1)^2-(sin(x)/x-1)^3+... =3-sin(x)/x + sin^2(x)/x^2- 2sin(x)/x - (sin(x)/x-1)^3+... = 3 - 3 sin(x)/x + sin^2(x)/x^2-...$

I am not sure how to continue from here. How can I plug in the series of sin(x)/x into the series of 1/(y+1) without everything blowing up immediately? And what do I do with the terms $sin^2(x)/x^2$ and those of higher order, where I am basically supposed to calculate the square of an infinite sum?
Thanks for your help though, I feel much closer to an answer already!

 

[Orodruin]

As already stated, you do not need to keep the entire infinite sum as you are only interested in the first terms. You can write (for example)
$$
\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6).
$$
The $\mathscr O$ will help you keep track of the order of the terms you have neglected.

 

[RedDwarf]

As already stated, you do not need to keep the entire infinite sum as you are only interested in the first terms. You can write (for example)
$$
\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6).
$$
The $\mathscr O$ will help you keep track of the order of the terms you have neglected.

Let's see if I can do this:
If $\sin(x)/x = 1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathscr O(x^6)$, then
$$1/(y+1) = 1-1 + \frac{x^2}{6} - \frac{x^4}{120} +1 + (-\frac{x^2}{6} + \frac{x^4}{120})^2
= 1+\frac{x^2}{6} - \frac{x^4}{120} +\frac{x^4}{36} - 2\frac{x^6}{6*120}+ \frac{x^8}{120})^2$$
Neglecting all terms with $x^6$ or higher:
$$x/\sin(x) = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \mathscr O(x^6)$$
Yeah, that looks like what I have seen before. Thanks a million!

 

[Orodruin]

You really should be writing out the $\mathscr O$ in every step or your equalities will not be correct. I also would not bother computing the higher order terms explicitly as they will be eaten by the $\mathscr O$ anyway and it already contains unknown coefficients. Just shove them into it right away. The general approach is valid though.