Question

Find a MacLaurin Series in table 1 to obtain a series for f(x)= (x-sinx)/x^3 . Write the final answer in sigma notation?

Answer 1

`f(x) = (x-sinx)/x^3 = sum_(k=0)^oo (-1)^k x^(2k)/((2k+3)!)`

Explanation:

I do not see any Table 1, anyway you can start from the MacLaurin series for `sinx`:

`sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)`

Extract from the sum the term for `n=0`:

`sinx = x + sum_(n=1)^oo (-1)^n x^(2n+1)/((2n+1)!)`

So:

`x-sinx = x - x -sum_(n=1)^oo (-1)^n x^(2n+1)/((2n+1)!) = sum_(n=1)^oo (-1)^(n+1) x^(2n+1)/((2n+1)!)`

and dividing by `x^3` term by term:

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

We can express the series with an index starting from zero by substituting `k=n-1`, so that:

`(-1)^(n+1) = (-1)^(k+2)= (-1)^k`

`x^(2n-2) = x^(2(n-1)) = x^(2k)`

`(2n+1)! = (2(n-1)+3)! = (2k+3)!`

so that:

`f(x) = (x-sinx)/x^3 = sum_(k=0)^oo (-1)^k x^(2k)/((2k+3)!)`