Prove the equality (Taylor series).

Question

Prove the equality: $$ \frac{1}{3}\left(e^x+2e^{-x/2}\cos\frac{x\sqrt{3}}{2}\right)= \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!},\ \ -\infty<x<+\infty $$

I tried to apply Euler's formula ($e^{ix}=\cos x+i\sin x$) to this problem but it went rather unsuccessful. Here is what I did: $$ e^{-x/2}=e^{i(ix/2)}=\cos\frac{ix}{2}+i\sin\frac{ix}{2}\Rightarrow\\ \Rightarrow 2e^{-x/2}\cos\frac{x\sqrt{3}}{2}=2\cos\frac{ix}{2}\cos\frac{x\sqrt{3}}{2}+ 2i\sin\frac{ix}{2}\cos\frac{x\sqrt{3}}{2}=\\ =\cos\frac{x(i+\sqrt{3})}{2}+\cos\frac{x(i-\sqrt{3})}{2}+ i\sin\frac{x(i+\sqrt{3})}{2}+i\sin\frac{x(i-\sqrt{3})}{2}=\\ =e^{ix(i+\sqrt{3})/2}+e^{ix(i-\sqrt{3})/2}= e^{x(-1+i\sqrt{3})/2}+e^{x(-1-i\sqrt{3})/2} $$ Then I tried to use Maclaurin series for $e^{x(-1+i\sqrt{3})/2}$ and $e^{x(-1-i\sqrt{3})/2}$ after which I got completely befuddled because it seemed to me that I had only complicated the initial problem.

So, if anyone could help me, I would appreciate it.

Answer 1

Hint: (A followup to Lord Shark the Unknown's observation) You're already half-way there. You've established that \begin{align} \frac{1}{3}\left(e^x + 2e^\frac{-x}{2}\cos \frac{x\sqrt{3}}{2}\right) &=\frac{1}{3}\left(e^x + e^\frac{x\left(-1+i\sqrt{3}\right)}{2}+ e^\frac{x\left(-1-i\sqrt{3}\right)}{2}\right)\\ &= \frac{1}{3}\left(e^{z_1 x} + e^{z_2x} + e^{z_3}x \right)\ , \end{align} where $\ z_1=1\ $, $\ z_2=\frac{x\left(-1+i\sqrt{3}\right)}{2}\ $, and $\ z_3=\frac{\left(-1-i\sqrt{3}\right)}{2}\ $ are the three cube roots of unity. If you now use the expansions \begin{align} e^{z_ix}&= \sum_{n=0}^\infty\frac{z_i^nx^n}{n!}\\ &=\sum_{n=0}^\infty\left(\frac{z_i^{3n}x^{3n}}{(3n)!}+\frac{z_i^{3n+1}x^{3n+1}}{(3n+1)!}+\frac{z_i^{3n+2}x^{3n+2}}{(3n+2)!}\right)\\ &=\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}\left(1 +\frac{z_ix}{3n+1}+\frac{z_i^2x^2}{3n+2}\right)\ , \end{align} and the observations that $\ z_1^2=z_2\ $, $\ z_2^2=z_1\ $, and $\ z_1 + z_2 + z_3=0\ $, you should be able to complete the demonstration.

Answer 2

With $~\displaystyle m\in\mathbb{N},~k\in\{0,1,2,…,m-1\}~$ and $~\lambda:=e^{i\frac{2\pi}{m}}~$

using series expansion of $~e^x~$ we get:

$$\sum\limits_{n=0}^\infty\frac{x^{mn+k}}{(mn+k)!}=\frac{1}{m}\sum\limits_{j=0}^{m-1}\lambda^{-kj}e^{x\lambda^j}$$

This is because of $~\lambda^{mn}|_{n\in\mathbb{Z}}=1~$ and $~\sum\limits_{j=0}^{m-1}\lambda^{kj}=m|_{k\equiv 0\,(mod\,m)} \lor 0|_{k\not\equiv 0\,(mod\,m)} ~$ .

Separating real and imaginary parts leads to the formula:

$\Re(\lambda^{-kj}e^{x\lambda^j})=\Re(\lambda^{-kj})\Re(e^{x\lambda^j})-\Im(\lambda^{-kj})\Im(e^{x\lambda^j})=$

$\hspace{2.4cm}=\cos\frac{2\pi kj}{m}e^{x\cos\frac{2\pi j}{m}}\cos\left(x\sin\frac{2\pi j}{m}\right)+\sin\frac{2\pi kj}{m}e^{x\cos\frac{2\pi j}{m}}\sin\left(x\sin\frac{2\pi j}{m}\right)$

$k=0~:~~\sin\frac{2\pi kj}{m}=0$

Now the special case $~(m,k):=(3,0)~$:

$$\sum\limits_{n=0}^\infty\frac{x^{3n}}{(3n)!}=\sum\limits_{j=0}^2 e^{x\cos\frac{2\pi j}{3}}\cos\left(x\sin\frac{2\pi j}{3}\right)=\frac{1}{3}\left(e^x+2e^{-\frac{x}{2}}\cos\left(\frac{x}{2}\sqrt{3}\right)\right)$$