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

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3 Answers

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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.

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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
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@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
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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.

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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 $k=3$.

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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
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