Maclaurin series for $\dfrac{\cos(x)}{x^2}$ I know how to solve maclaurin series for $\cos(x)$, but for $\dfrac{\cos(x)}{x^2}$ do i have to divide each term by $x^2$?
 A: That is a generalized expansion.
$$\frac{\cos(x)}{x^2}=\frac{1}{x^2}+\sum_{k=1}^{+\infty}\frac{(-1)^kx^{2k-2}}{(2k)!}$$
A: For a Maclaurin expansion to exist and converge all the derivatives of the functions must exist at $x=0$. 
But we can show that $\frac{\cos(x)}{x^2}$ has neither a 1st nor 0th order derivative at $x=0$. 
However it is possible to show the function has a Laurent series around $x=0$, with a radius of convergence that is squeezed in between some two positive real numbers.
A: Due to the asymptote at the origin, the function has no McLaurin development. Anyway, you can take the singularity away with
$$\frac{\cos x}{x^2}=\frac{\cos x-1}{x^2}+\frac1{x^2}.$$
Now you indeed obtain the McLaurin development of the non-singular part using the development of the cosine and simplifying by $x^2$,
$$f(x)=\frac{\sum_{k=0}^\infty{(-1)^{k}\frac{x^{2k}}{(2k)!}-1}}{x^2}=\sum_{k=1}^\infty{(-1)^{k}\frac{x^{2k-2}}{(2k)!}}.$$
The same development can be obtained by direct evaluation of the derivatives, in a much more tedious way:
$$f(0)=\lim_{x\to0}\frac{\cos x-1}{x^2}=\lim_{x\to0}\frac{-\sin x}{2x}=-\frac12.$$
$$f'(0)=\lim_{x\to0}\frac{-x\sin x-2\cos x+2}{x^3}=\lim_{x\to0}\frac{-\sin x-x\cos x+2\sin x}{3x^2}\\
=\lim_{x\to0}\frac{-\cos x+\cos x-x\sin x+2\cos x}{6x}=0$$
$$\cdots$$
A: Hint:
$$\cos(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$$
Try dividing both sides by $x^2$.
