How do I find $S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2 p}}{(2 p) !}+\frac{x^{3 p}}{(3 p) !}+\cdots$ by complex numbers, where $p \in N$?

Solution 1:

We are going to evaluate the series $\displaystyle S(p)=1+\frac{x^{p}}{p !}+\frac{x^{2p}}{(2p) !}+\frac{x^{3p}}{(3p )!}+\cdots $

Let’s start with the series $ \displaystyle e^{z}=\sum_{x=0}^{\infty} \frac{z^{k}}{k !} \tag*{(*)} $ Now consider the complex $p^{th} $ root of unity $\omega=e^{\frac{ 2\pi}{p}i} $ satisfying $$ 1+\omega+\omega^{2}+\cdots+\omega^{p-1} =0 \text{ and }\omega^{p}=1.$$ Putting $ z=x$ yields $ \displaystyle e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} \tag*{(1)}$ Putting $ z=\omega x $ yields $ \displaystyle e^{\omega x}=\sum_{k=0}^{\infty} \frac{\omega^{k} x^{k}}{k !}\tag*{(2)}$ Putting $\displaystyle z=\omega^{2} x $ yields $ \displaystyle e^{\omega^{2} x}=\sum_{k=0}^{\infty} \frac{\omega^{2 k} x^{k}}{k !} \tag*{(3)}$ $$\vdots$$

Putting $\displaystyle z=\omega^{p-1} x $ yields $ \displaystyle e^{\omega^{p-1} x}=\sum_{k=0}^{\infty} \frac{\omega^{(p-1)k} x^{k}}{k !} \tag*{(p-1)}$

$(1)+(2)+(3)+\cdots +(p-1)$ gives $ \displaystyle \sum_{k=0}^{\infty} \frac{1+\omega^{k}+\omega^{2 k}+\cdots+\omega^{(p-1)k}}{k !}x^{k}=e^{x}+e^{\omega x}+e^{\omega^{2} x}+ \cdots +e^{\omega^{p-1} x}\tag*{}$ $ \begin{aligned} \displaystyle \because 1+\omega^{k}+\omega^{2 k}+\cdots+ \omega^{(p-1) k} &=\left\{\begin{array}{cl}\frac{1-\left(\omega^{p}\right)^{k}}{1-\omega^{p}} & \text { if } p\not | k \\ p & \text { if } p|k\end{array}\right. =\left\{\begin{array}{ll}0 & \text { if } p\not|k \\p & \text { if } p | k\end{array}\right.\\\displaystyle \therefore \sum_{k=0}^{\infty} \frac{p}{(p k) !} x^{p k} &=e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \end{aligned} \tag*{} $

$$ S(p)= \sum_{k=0}^{\infty} \frac{x^{p k}}{(p k) !} =\frac{1}{p} \left(e^{x}+e^{\omega x}+e^{\omega^{2} x} +\cdots + e^{\omega^{p-1} x} \right)\tag*{(**)} $$ When $p$ is odd,

$$ \begin{aligned} S(p) &=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\ &=\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}}\left(e^{x e^{\frac{2 k \pi}{p}i} }+e^{x e^{\frac{2(p-k) \pi}{p} i}}\right)\right]\\& =\frac{1}{p}\left[e^{x}+\sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(e^{i x \sin \frac{2k\pi}{p}}+e^{-i x \sin \frac{2k \pi)}{p}}\right)\right]\\ &=\frac{1}{p}\left[e^{x}+2 \sum_{k=1}^{\frac{p-1}{2}} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right] \end{aligned} $$ When $p$ is even,

$$\begin{aligned} S(p) &=\frac{1}{p}\left[e^{x}+e^{-x}+\sum_{k=1}^{\frac{p}{2} -1}\left(e^{\omega^{k} x}+e^{\omega^{p-k} x}\right)\right]\\ &=\frac{1}{p}\left[e^{x} +e^{-x} +2 \sum_{k=1}^{\frac{p}{2}-1} e^{x \cos \frac{2 k x}{p}}\left(\cos \left(x \sin \frac{2 k \pi}{p}\right)\right)\right] \end{aligned}$$

By the way, by successive differentiation on $S(p)$, we can find, for any natural number $n<p$, the series, $$ \frac{x^{n}}{n!}+\frac{x^{n+p}}{(n+p) !}+\frac{x^{n+2 p}}{(n+2 p) !}+\cdots $$ whose length is $p$ and starting terms other than $1$.