Function representing a Taylor Series

Question

Find a function represented by the Taylor series $\sum_{k=0}^{\inf}\left(-1\right)^{k}\left(\frac{3^{2k+1}}{\left(2k+1\right)!}\right)x^{2k}$.

After taking a bunch of derivatives, I figured out $f^n(x)=\frac{3^{2n+1}}{2n+1}$, which further simplified to $\frac{3(3x)^2k}{(2k+1)!}$. I'm stuck here, I think I have to use $\sin(x)$ somehow, but I can't figure out how. Here's the graph. How should I continue?

Answer 1

Using the Taylor series for the Sine function as mentioned namely $$\sin{(x)}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}$$ we have that your sum (provided at the link) is simply $$\frac1{x}\sum_{k=1}^\infty\frac{(-1)^k(3x)^{2k+1}}{(2k+1)!}=\frac{\sin{(3x)}}{x}-3$$

Answer 2

Hint:

The series for the sine is

$$\sin x=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}.$$

Now,

$$(3^{2k}+1)x^{2k}=\frac{(3x)^{2k+1}}{3x}+\frac{x^{2k+1}}x.$$ The function can be written as

$$\frac13\text{sinc}(3x)+\text{sinc}(x)$$ where $\text{sinc}$ is called the cardinal sine.