Expand a sin(x^3) in Maclaurin's series and find a 30th derivative at (0)

Question

I have a big task and problems with it.

I have to expand this function in Maclaurin's series. $$cos(x^{3})$$ I tried expand it as $$\sum_{n=0}^\infty (x^n)/n!$$ but it's for $cos(x)$. So i don't know how to expand and rewrite $cos(x^3)$ in Maclaurin's series. Can you tell me how to find 30th derivative of this function

I would really appreciate your help

Answer 1

You know that

$$\cos y = 1 - \frac{y^2}{2!} + \frac{y^4}{4!} - \dots$$

So you substitute $y = x^3$ to get

$$\cos x^3= 1 - \frac{x^6}{2!} + \frac{x^{12}}{4!} - \dots - \frac{x^{30}}{10!} + \dots$$

If you differentiated both sides $30$ times, the constant term on the right will be $-\frac{30!}{10!}$. All following terms will be dependant on $x$. Substituting $x = 0$ to find the 30th derivative will cause them all to evaluate to $0$.

Answer 2

Observe that $$ sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}. $$ Plugging in $x^3$ then gives us $$ sin(x^3)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(x^3)^{2n+1}}{(2n+1)!}. $$ Differentiate this a few times with respect to $x$ to see the pattern for the 30th derivative.

Answer 3

Hint When evaluated at $x = 0$, the only term of the series which is non-zero is the constant.

Hint The constant term of $f^{(30)} (x)$ is equal to the coefficient of $x^{30}$ in $f(x)$ multiplied by $30!$