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

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

• It should...but it musn't necessarily. Nov 28 '18 at 8:22
• I think you could take derivatives term by term [maybe several times] and obtain differential equations about this function. Then solve this ODE.
– xbh
Nov 28 '18 at 8:23
• Nov 28 '18 at 9:02
• Sorry, didn't notice :s Nov 28 '18 at 9:52

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

• Very elegant answer, thank you^^ Nov 28 '18 at 9:28
• @SADBOYS You are welcome! I started differentianting and then noticed that :). Of course one need to observe that the series converges for any $x$ in order to consider $f(x)$.
– user
Nov 28 '18 at 9:32
• Gimusi. Very nice!! Nov 28 '18 at 10:34
• @PeterSzilas Thanks my friend! :)
– user
Nov 28 '18 at 10:42

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

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

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

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

• Nov 28 '18 at 8:27
• As we need every third we have use $1+w+w^2=0$ Can you generalize ? Nov 28 '18 at 8:28