Calculate $\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$

Question

$$\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$ 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 1

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 $f(0)=1$, $f'(0)=0$, $f''(0)=0$.

Answer 2

$$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

Put $y=x,xw,xw^2$ where $w$ is a complex cube root of unity

Now if $w=\dfrac{-1+\sqrt3i}2,w^2=\dfrac{-1-\sqrt3i}2$,

$$e^x+e^{wx}+e^{w^2x}=e^x+e^{-x/2}\left(e^{\sqrt3ix/2}+e^{-\sqrt3ix/2}\right)=?$$