3
$\begingroup$

How do I find the Maclaurin Series for $x^3 \sin{2x}$? If I start differenciating, I get 2 terms like $2x^3 \cos{2x} + \sin{2x}\cdot 3x^2$ then 4 for the next one. Is this the right way to go?

I just need to find $f^{(2012)}(0)$ of $f(x)= x^3\cdot \sin{2x}$

3 Answers 3

2

You know that $\sin x=\sum_{n\ge 0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\;;\tag{1}$ to get $\sin 2x$, just substitute $2x$ for $x$ in $(1)$, and to get $x^3\sin 2x$, follow that up by multiplying by $x^3$. Then you need only figure out what the coefficient of $x^{2012}$ is.

  • 1
    @JiewMeng: The $x^{2012}$ term is indeed the one for which $n=1004$: it’s $\frac{(-1)^{1004}}{2009!}x^{2012}\;,$ so the coefficient of $x^{2012}$ is $\frac1{2009!}\;.$ The coefficient of $x^{2012}$ in the Maclaurin series for $f$ is $\frac{f^{(2012)}(0)}{2012!}\;.$ When you equate these, what do you get for $f^{(2012)}(0)$?2012-03-19
1

Since this is homework, just a few hints:

  • Find the series $T(x)$ for $\sin x$ around $x=0$.

  • Evaluate $T(x)$ at $2x$ and simplify so it looks like another Taylor/Maclaurin series.

  • Multiply $T(2x)$ by $x^3$ to obtain the series for $x^3\sin(2x)$.

For more examples, see this Planetmath page.

0

If you just want to find $f^{(2012)}(0)$, then a good choice may be Leibniz rule for differentiating a product $n$ times. Note that your formula will end at $k=3$.

  • 0
    Thanks @Brian M. Scott for fixing the link.2012-03-19