Question

Use Maclauin series of

E^z to compute
series from 0 to infinity of
[Cos (n.phi/3) ] / [ n!]

Answer 1

Answer on Question#49959 - Math -Complex Analysis.

Use Maclauin series of $e^z$ to compute series from 0 to $\infty$ of $\frac{\cos\left(\frac{n\varphi}{3}\right)}{n!}$.

Solution. Let $\varphi$ be real and $z = e^{i\varphi /3}$. Maclaurin for $e^z$ is $e^z = \sum_{k = 0}^{\infty}\frac{z^k}{k!}$. Our sum is

$$\sum_{k = 0}^{\infty}\frac{\cos\left(\frac{n\varphi}{3}\right)}{n!} = \sum_{k = 0}^{\infty}Re\frac{z^n}{n!} = Re\sum_{k = 0}^{\infty}\frac{z^n}{n!} = Re(e^z) = Re\left(e^{\cos\left(\frac{\varphi}{3}\right) + \operatorname{isin}\left(\frac{\varphi}{3}\right)}\right) = 0$$ $$Re\left(e^{\cos\left(\frac{\varphi}{3}\right)}\left(\cos\left(\sin\left(\frac{\varphi}{3}\right) + \operatorname{isin}\left(\frac{\varphi}{3}\right)\right)\right) = e^{\cos\left(\frac{\varphi}{3}\right)}\cos\left(\sin\left(\frac{\varphi}{3}\right)\right)\right).$$

Answer: $e^{\cos\left(\frac{\varphi}{3}\right)}\cos\left(\sin\left(\frac{\varphi}{3}\right)\right)$.

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