# Show that $\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!} = \frac{1}{3} e^x + \frac{2}{3} e^{-\frac{x}{2}} \cos\left(\frac{\sqrt{3}}{2} x\right)$

Is there anyone who knows, and want to help, how to show that this is true $\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!} = \frac{1}{3} e^x + \frac{2}{3} e^{-\frac{x}{2}} \cos\left(\frac{\sqrt{3}}{2} x\right)$ ?

I know that $\sum_{k=0}^{\infty}\frac{x^{k}}{k!}= e^x$, but how to use it I don't know. I can't find my lecture notes so it is like that.

• Use $\cos y=\frac12(e^y+e^{-y})$. – Angina Seng Jan 5 '18 at 16:55
• @LordSharktheUnknown you forgot the $i$'s – Shashi Jan 5 '18 at 16:56
• @LordSharktheUnknown that looks more like $\cosh(x)$ – Henry Jan 5 '18 at 16:57

Outline: Differentiate three times and notice that the series satisfies $y'''=y$. Then solve that differential equation, and match with the conditions $y(0)=1$ and $y'(0)=y''(0)=0$, that you get by looking at the coefficients of the series.
• $\sum_{k=0}^{\infty}[\frac{1}{x^3 (3k-3)!}-\frac{1}{(3k)!} ]x^{3k} -1 = 0$ is this in the right direction? This is what I get when trying to solve y'''-y=0 – 123 Jan 5 '18 at 17:30
• @potete I think using the characteristic polynomial will be better than that. Just solve $\lambda^3-1=0$ you will get three different lambda's and your solution will be $$y=Ae^{\lambda_1 x}+Be^{\lambda_2 x} +Ce^{\lambda_3 x}$$ don't forget to make your solution real. Using series is circular. – Shashi Jan 5 '18 at 17:45
Outline: Let $\omega = \exp(2\pi i/3) = -1/2 + i\sqrt3/2$. Apply the exponential series to $\exp(x) + \exp(\omega x) + \exp(\omega^2x)$. Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, the result is $3$ times the series you're interested in.