# Finding the Maclaurin series for $x^2 \sin(x^3)$

I have a question here;

suppose $f(x)= x^2\sin(x^3)$

By using the Maclaurin series for sine, find the Maclaurin series for $f$

I understand how to obtain the Maclaurin series for $f$ using the Maclaurin series for $\cos(x)$ - by substituting $x^3$, and then finding the derivative and adjusting the scalar multiplier - but how am I able to do this using sin to answer the question?

Should I just find the Maclaurin series for $\cos(x)$ using the Maclaurin series I found for $\sin(x)$ and use that to justify having satisfied the question, or is there a more direct way to do this?

Thanks for any help on this

• The question asked you to use the Maclaurin series of $\sin(x)$. So just use that series and substitute $x^3$ into each $x$ term. Nov 3, 2015 at 3:54
• but how to get the x^2 out the front? Nov 3, 2015 at 3:59
• you can use product rule to differentiate it and obtain all the higher order derivatives. after that just plug 0. Nov 3, 2015 at 4:00
• First find the Maclaurin series of $\sin(x^3)$. Then multiply $x^2$ to each term of the series. Nov 3, 2015 at 4:00
• ah multiplying by x^2 seems so obvious now. I'm new to this stuff, I'm used to thinking of x as the variable Nov 3, 2015 at 4:02

The Maclaurin expansion for $\sin(x)$ is $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}. \tag{1}$$ In order to get the Maclaurin expansion for $\sin(x^3)$, we plug in $x^3$ into each $x$ appearing in $(1)$, $$\sin(x^3) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \cdots = \sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{(2n+1)!}.$$ And finally, in order to go from $\sin(x^3)$ to $x^2 \sin(x^3)$, we just multiply each term by $x^2$, $$x^2 \sin(x^3) = x^5 - \frac{x^{11}}{3!} + \frac{x^{17}}{5!} - \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{6n+5}}{(2n+1)!}.$$ This is the desired expansion. $\diamondsuit$