Question

Can someone help me find the Maclaurin series for the given function?

I am trying to find the Maclaurin series for this function: f(x) = x^5 cos(πx). I am having trouble because the derivatives get large fast and I'm not sure if I'm doing it right.

Answer 1

`x^5cospix = sum_(n=0)^oo \ (-1)^n (pi^(2n) x^(2n+5))/(2n!)`

`\ \ \ \ \ \ \ = 1 - (pi^2 x^7)/(2) + (pi^4 x^9)/(24) - (pi^6 x^11)/(720) + ...`

Explanation:

We seek a Maclaurin series for

`f(x) = x^5cos(pix)`

We start by using the well studied Maclaurin series for `cos x`:

`cosx = sum_(n=0)^oo \ (-1)^n x^(2n)/(2n!)`
`\ \ \ \ \ \ \ = 1 - x^2/(2!) + x^4/(4!) - x^6/(6!) + x^8/(8!) + ...`

So that:

`cospix = sum_(n=0)^oo \ (-1)^n (pi x)^(2n)/(2n!)`
`\ \ \ \ \ \ \ = 1 - (pix)^2/(2!) + (pix)^4/(4!) - (pix)^6/(6!) + (pix)^8/(8!) + ...`

And finally:

`x^5cospix = sum_(n=0)^oo \ (-1)^n x^5(pi x)^(2n)/(2n!)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = sum_(n=0)^oo \ (-1)^n (x^5 \ pi^(2n) x^(2n))/(2n!)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = sum_(n=0)^oo \ (-1)^n (pi^(2n) x^(2n+5))/(2n!)`

`\ \ \ \ \ \ \ = 1 - (pi^2 x^7)/(2) + (pi^4 x^9)/(24) - (pi^6 x^11)/(720) + ...`