# Taylor expansion of x/sin(ax)

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.

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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
Staff Emeritus
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.