Question

How do you find the Maclaurin Series for `(sinx-x)/x^3`?

Answer 1

`(sinx-x)/x^3 = sum_{k=1}^oo(-1)^k (x^{2k-2})/((2k+1)!)`

Explanation:

The series expansion for `sin x` is given for

`sin x = sum_{k=0}^oo(-1)^k (x^{2k+1})/((2k+1)!) =`
`= x -1/(3!)x^3+1/(5!) x^5 -1/(7!) x^7 + cdots`

then

`(sinx-x)/x^3 = sum_{k=1}^oo(-1)^k (x^{2k-2})/((2k+1)!)`