# How to calculate the Maclaurin series for $\frac{{x - \sin (x)}}{{{x^2}}}$

For the function $$f(x)=\frac{{x - \sin (x)}}{{{x^2}}},\quad x\neq 0$$ it is known that the Maclaurin series of the function $f(x)$ of class $C^\infty$ is equal to the corresponding Taylor series $$\sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(c)}}{{n!}}{{(x - c)}^n}}$$ with $c=0$ parameter. But $f(c)$ and $f^{(n)}(c)$ are not defined for $c=0$.

What would be the Maclaurin series?

They can give me some suggestions please?

• Do you know the series for $\sin(x)$? – Andrés E. Caicedo Dec 2 '13 at 6:42
• yes, but how doing for the series $\frac{{\sin (x)}}{{{x^2}}}$? – mathsalomon Dec 2 '13 at 6:47
• @mathsalomon Do you know the series for $x-\sin{x}$? – David H Dec 2 '13 at 6:54
• @DavidH Thanks. You have given me a good idea – mathsalomon Dec 2 '13 at 6:57

$\sin x=\sum_{\color{red}{k=0}}^{\infty} \frac{(-1)^{\color{red}{k}}}{(2k+1)!}x^{2k+1}$ , so $x-\sin x=\sum_{\color{red}{k=1}}^{\infty} \frac{(-1)^{\color{red}{k+1}}}{(2k+1)!}x^{2k+1}$ and finally $$\frac{x-\sin x}{x^2}=\sum_{\color{red}{k=1}}^{\infty} \frac{(-1)^{\color{red}{k+1}}}{(2k+1)!}x^{\color{red}{2k-1}}$$
Take the series at $x=0$ for $\sin(x)$, which starts $0+x+\cdots$, subtract from$~x$ (after which there are no terms of degree $0$ or $1$, so everything that remains is divisible by $x^2$ at least), and divide out the $x^2$.
• @mathsalomon: Or above (depending on your preferences). It happens that my answer was not posted for about a quarter of an hour due to internet connection problems, so I did not really see the other answer. Anyway, I think the main point is that you can compute Maclaurin series for all ingredients of the formula (here essentially for $\sin(x)$) and after that operate on the formal series. In fact Maclaurin series are formal series: they doe not necessarily converge at any $x\neq0$, although in the most commonly seen cases (like this one) they do converge. – Marc van Leeuwen Dec 2 '13 at 7:35