Question

Find the first four nonzero terms of the taylor series about 0 for the function f(x)=x4sin(6x). note that you may want to find these in a manner other than by direct differentiation of the function.

Answer 1

The quickest way to get the series is if you already know the series for sine:

$\sin6x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n(6x)^{2n+1}}{(2n+1)!}$

Then just multiply the series by $x^4$ to get

$f(x)=x^4\sin6x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n6^{2n+1}x^{2n+5}}{(2n+1)!}$

The first four nonzero terms are simply the first four terms:

$6x^5-\dfrac{6^3}{3!}x^7+\dfrac{6^5}{5!}x^9-\dfrac{6^7}{7!}x^{11}$