Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f(x)= x^3\cdot \sin{2x}$

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
Hmm ... so $x^3 \sin{2x} = \sum \frac{(-1)^n (2x)^{2n+1}(x^3)}{(2n+1)!}$? If so why don't I need to differenciate $x^3$ like what I need before I get the maclaurin series for $\sin$? – Jiew Meng Mar 18 '12 at 16:14
@JiewMeng: That’s right; and you can simplify $(2x)^{2n+1}x^3$ to $2^{2n+1}$ times what power of $x$? – Brian M. Scott Mar 18 '12 at 16:15
I got $x^3 \sin{2x} = \sum \frac{(-1)^n 2^{2n+1} x^2n+4}{(2n+1)!}$. How do I get rid of the $x^{2n+4}$ before I find $f^{(2012)}(0)$? In a prev problem I canceled it with $x^n$. Else I will have $f^{(n)}(0) = \frac{(-1)^n 2^{2n+1} x^{2+4/n} n!}{(2n+1)!}$ – Jiew Meng Mar 18 '12 at 16:24
@JiewMeng: You want the coefficient of $x^{2012}$. For what value of $n$ is $2n+4=2012$? What is the coefficient of that term? You’ll never have powers of $x$ in $f^{(n)}(0)$: that’s just a number. – Brian M. Scott Mar 18 '12 at 16:25
@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 $f^{(2012)}(0)$? – Brian M. Scott Mar 19 '12 at 2:00

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.

share|cite|improve this answer

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 $k=3$.

share|cite|improve this answer
Your answer will be something like $3!\binom{2012}{3}(\sin 2x)^{(2009)}(0)=..$ – Tapu Mar 18 '12 at 18:33
Thanks @Brian M. Scott for fixing the link. – Tapu Mar 19 '12 at 21:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.