How do you find the Maclaurin series for x^3cos(x^2)?

1 Answer
Jul 12, 2016

x^3 cos(x^2) = sum_(n=0)^(∞) (-1)^(n) (x^(4n+3))/((2n)!) = x^3-(x^7)/(2!)+(x^11)/(4!)-(x^15)/(6!) + ...

Explanation:

To find a Maclaurin series for x^3 cos(x^2), we first need to take note of some very well-known series, such as

cos x = sum_(n=0)^(∞) (-1)^(n) (x^(2n))/((2n)!)

We now have to find a Maclaurin series for cos(x^2), for which all we need to do is replace x with x^2 in the following way:

cos(x^2) = sum_(n=0)^(∞) (-1)^(n) (x^(2))^(2n)/((2n)!) = sum_(n=0)^(∞) (-1)^(n) (x^(4n))/((2n)!)

Multiplying out our x^3 term, we get

x^3 cos(x^2) = x^3 * sum_(n=0)^(∞) (-1)^(n) (x^(4n))/((2n)!) = sum_(n=0)^(∞) (-1)^(n) (x^(4n+3))/((2n)!)

Finally, to test this Maclaurin series to roughly 4 terms, we can write

x^3 cos(x^2) = sum_(n=0)^(∞) (-1)^(n) (x^(4n+3))/((2n)!) = x^3-(x^7)/(2!)+(x^11)/(4!)-(x^15)/(6!) + ...

To see whether this approximation to 4 terms is accurate:

Graph of x^3 cos(x^2)
graph{x^3 cos(x^2) [-10, 10, -5, 5]}
Graph of x^3-(x^7)/(2!)+(x^11)/(4!)-(x^15)/(6!)
graph{x^3-(x^7)/(2!)+(x^11)/(4!)-(x^15)/(6!) [-10, 10, -5, 5]}