The first 4 terms of the Maclaurin series for $\frac{x}{\sin x}$?

Question

I have to find the first 4 terms of $\frac{x}{\sin x}$. My first step was to calculate the first 4 terms of the Maclaurin series for $\sin x$ which are $$\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}$$ Then, we will have that the first 4 terms of $\frac{x}{\sin x}$ are: $$ \frac{x}{\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}}$$ which would give me $$1 - \frac{3!}{x^2} + \frac{5!}{x^4} - \frac{7!}{x^6}$$ however that answer is incorrect. My question is: why can't we just divide by $x$ to get the answer? When we do the Maclaurin series for $\frac{\sin x}{x}$ we just divide by $x$ (see attached picture below)

Why does this same concept not work when getting the Maclaurin series for $\frac{x}{\sin x}$?

Answer 1

Notice that when you divide by more one term you can not divide term by term.

For example $$ \frac {1}{2+3} \ne \frac {1}{2}+\frac {1}{3}$$

Thus dividing $x$ by $\sin x$ requires more than just dividing term by term.

You may assume that $$\frac {x}{\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+...} =a_0+a_1x +a_2x^2 +...$$ and try to find the coefficients by cross multiplication.

Answer 2

The problem is $\frac{1}{\sum_ka_kx^k}=\sum_k\frac{1}{a_kx^k}$ isn't an identity. Write $\frac{x}{\sin x}=1+ax^2+bx^4+cx^6+o(x^6)$ so$$1+o(x^6)=(1+ax^2+bx^4+cx^6)\left(1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6\right).$$The $x^2$ coefficient gives $a=\frac16$; the $x^4$ coefficient gives $b=\frac{a}{3!}-\frac{1}{5!}=\frac{7}{360}$; the $x^6$ coefficient gives$$c=\frac{b}{3!}-\frac{a}{5!}+\frac{1}{7!}=\frac{31}{15120}.$$See also here.

Answer 3

Begin with the expansion you proposed:

$$\frac{x}{\sin{x}} = \frac1{1-(x^2/3! - x^4/5! + x^6/7! -...)}$$

Because we are expanding about $x=0$ we can treat the terms in parentheses as a term to be expanded in a geometric series. The expansion then takes the form

$$1 + ()+ ()^2 + ()^3 +...$$

Note that if we want four terms then we are expanding out to $O(x^6)$. Accordingly we need only worry about the $()^3$ and lower terms. Therefore the expansion sought is

$$\frac{x}{\sin{x}} = 1 + \frac{x^2}{6} + \frac{7 x^4}{360} + \frac{31 x^6}{15120} + O(x^8)$$

Answer 4

It is just need to know the series expansion \begin{equation}\label{csc-ser-eq}\tag{CSCSEREX} \csc x=\frac{1}{\sin x}=\frac1x+\sum_{k=1}^\infty\frac{2\bigl(2^{2k-1}-1\bigr)|B_{2k}|}{(2k)!}x^{2k-1}, \quad |x|<\pi, \end{equation} where the Bernoulli numbers $B_k$ are generated by \begin{equation*} \frac{z}{\text{e}^z-1}=\sum_{k=0}^\infty B_k\frac{z^k}{k!}=1-\frac{z}2+\sum_{k=1}^\infty B_{2k}\frac{z^{2k}}{(2k)!}, \quad |z|<2\pi. \end{equation*} The series expansion \eqref{csc-ser-eq} can be looked up in almost any mathematical handbook.