power series - Calculate $sumlimits_{n=0}^{infty} frac{x^{3n}}{(3n)!}$

$$\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$ should be calculated using complex numbers I think, the Wolfram answer is :

$\frac{1}{3} (e^x + 2 e^{-x/2} \cos(\frac{\sqrt{3}x}{2}))$

How to approach this problem?

Answer

We have that by $f(x)=\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$

$$f'(x)=\frac{d}{dx}\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}$$

$$f''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}$$

$$f'''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-3}}{(3n-3)!}=f(x)$$

and $f'''(x)=f(x)$ has solution

$$f(x)=c_1e^x+c_2e^{-x/2}\cos\left(\frac{\sqrt 3 x}{2}\right)+c_3e^{-x/2}\sin\left(\frac{\sqrt 3 x}{2}\right)$$

with the initial conditions $f(0)=1$, $f'(0)=0$, $f''(0)=0$.