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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.

I'm thankful for your help.

Ángel Mario Gallegos
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123
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2 Answers2

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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.

Michael Behrend
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  • My friend, I don't see why you apply the first thing to exp(x)+exp(wx)+exp(w^2 x) – 123 Jan 05 '18 at 17:38
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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.

mickep
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  • $\sum_{k=0}^{\infty}[\frac{1}{x^3 (3k-3)!}-\frac{1}{(3k)!} ]x^{3k} -1 = 0$ is this in the right direction? This is what I get when trying to solve y'''-y=0 – 123 Jan 05 '18 at 17:30
  • If not then I try again. – 123 Jan 05 '18 at 17:32
  • You probably have solved that kind of differential equation before. It is linear and homogeneous. Use the method of characteristic equation. – mickep Jan 05 '18 at 17:41
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    @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 05 '18 at 17:45
  • You need to find the complex roots. – mickep Jan 05 '18 at 18:05