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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.

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  • $\begingroup$ $i(i+\sqrt3)/2=(-1+i\sqrt3)/2$ is a cube root of unity. It satisfies $z^3=1$. $\endgroup$ – Angina Seng Oct 27 '19 at 20:41
  • $\begingroup$ Look up multisection of series. $\endgroup$ – marty cohen Oct 27 '19 at 20:46
  • $\begingroup$ But how can it help (that $z^3=1$)? $\endgroup$ – Bonrey Oct 27 '19 at 20:47
  • $\begingroup$ You can easily show by studying the series as a power series that it converges uniformly on every bounded subset of R, so it can be differentiated term by term. At this point see if you can notice any pattern by differentiating it and seeing if it satisfies any particular differential equation, and see if you can solve said diff. eq. $\endgroup$ – user622002 Oct 27 '19 at 21:30
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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.

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  • $\begingroup$ Could you, please, explain how you got the last sum ($\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}\left(1 +\frac{z_ix}{3n+1}+\frac{z_i^2x^2}{3n+2}\right)$). It seems unobvious to me. $\endgroup$ – Bonrey Oct 28 '19 at 17:18
  • $\begingroup$ I've added an extra line to the expansion. Does that help? I still wouldn't categorise the step from the second to the third line as "obvious", since there's still a not-quite-trivial use of the observation $\ z_i^3=1\ $ that you need to make use of. $\endgroup$ – lonza leggiera Oct 28 '19 at 21:34
  • $\begingroup$ Jeez, I finally got your solution. It is really simple. I don't know why I was so stupid at first. Thank you! $\endgroup$ – Bonrey Oct 30 '19 at 7:45
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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 $~(m,k):=(3,0)~$:

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

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