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

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marked as duplicate by Martin R, Hans Lundmark, Arnaud D., Adrian Keister, Somos Nov 28 '18 at 13:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It should...but it musn't necessarily. $\endgroup$ – DonAntonio Nov 28 '18 at 8:22
  • $\begingroup$ I think you could take derivatives term by term [maybe several times] and obtain differential equations about this function. Then solve this ODE. $\endgroup$ – xbh Nov 28 '18 at 8:23
  • $\begingroup$ Related: math.stackexchange.com/q/686423, math.stackexchange.com/q/1193695. $\endgroup$ – Martin R Nov 28 '18 at 9:02
  • $\begingroup$ Sorry, didn't notice :s $\endgroup$ – SADBOYS Nov 28 '18 at 9:52
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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 $f(0)=1$, $f'(0)=0$, $f''(0)=0$.

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  • $\begingroup$ Very elegant answer, thank you^^ $\endgroup$ – SADBOYS Nov 28 '18 at 9:28
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    $\begingroup$ @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 $f(x)$. $\endgroup$ – gimusi Nov 28 '18 at 9:32
  • $\begingroup$ Gimusi. Very nice!! $\endgroup$ – Peter Szilas Nov 28 '18 at 10:34
  • $\begingroup$ @PeterSzilas Thanks my friend! :) $\endgroup$ – gimusi Nov 28 '18 at 10:42
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$$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

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

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

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

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