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]}
