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I have been learning about sums of series, and am very curious:

If we know that $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

What is the value of the following power series:

$S_1=1+\frac{x^3}{3!}+\frac{x^6}{6!}+...$ ,

$S_2=x+\frac{x^4}{4!}+\frac{x^7}{7!}+...$ ,

$S_3=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+...$ ?

PS: I know that if we divide $e^{ix}$ into two alternating term sums, we get $cos(x)$ and $isin(x)$ respectively. And for $e^x$, $cosh(x)$ and $sinh(x)$. Just trying to find out how it would work for three alternating term sums.

Researcher_Witty
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1 Answers1

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Denote $\omega$ as the cube-root of unity. We then have $1+\omega+\omega^2 = 0$. This gives us \begin{align} e^x & = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \cdots & \spadesuit\\ e^{\omega x} & = 1 + \dfrac{\omega x}{1!} + \dfrac{\omega^2 x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{\omega x^4}{4!} + \dfrac{\omega^2 x^5}{5!} + \cdots & \diamondsuit\\ e^{\omega^2 x} & = 1 + \dfrac{\omega^2 x}{1!} + \dfrac{\omega x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{\omega x^4}{4!} + \dfrac{\omega x^5}{5!} + \cdots & \clubsuit\\ \end{align} First series is given by $\dfrac{\spadesuit + \diamondsuit + \clubsuit}3$.

Second series is given by $\dfrac{\omega^2\spadesuit + \diamondsuit + \omega\clubsuit}{3\omega^2}$.

Third series is given by $\dfrac{\omega^2\spadesuit + \diamondsuit + \omega\clubsuit}{3\omega^2}$.

Adhvaitha
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