This question is a generalized form of the problem I asked before:
Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$
Let, look at this periodic sequence:
$$a_n=\left\{a_1,a_2,a_3,a_4,a_5,\cdots a_k ; a_1,a_2,a_3,a_4,a_5, \cdots a_k; a_1,a_2,a_3,a_4,a_5,\cdots a_k;\cdots \right\}$$, where $\left\{a_1, a_2, a_3, \cdots a_k \right\} \in \mathbb{Z^{+}}$ and $a_{k+1}=a_1, a_{k+2}=a_2, a_{k+3}=a_3, \cdots a_{2k}=a_k, \cdots$
For the sequence $a_k=\left\{ a_1,a_2,a_3,…,a_k \right\}$ , $k$ is a finite number. $a_1,a_2,…a_k$ are arbitary numbers.
Verbally, the series $a_n$ consists of an infinite number of periodic repetitions of the finite series $a_k.$
Finally my question is:
a) If there is an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?
b) If there is not an exist a algebraic closed form, for finite series $a_k$, in this case, does the $ a_n $ series always have a algebraic closed form?
I mean ,for example:
a)
$a_n=\left\{ 1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7,1,3,5,7\cdots\right\}$
b)
$a_n=\left\{1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9,1,8,2,6,5,9\cdots\right\}$
Thank you very much.
Best Answer
A periodic sequence with period $P$ can always be written as a trigonometric polynomial $$a_n = \sum_{j=0}^{P-1} b_j \cos(2 \pi j n/P) + \sum_{j=1}^{P-1} c_j \sin(2 \pi j n/P)$$ for some coefficients $b_j$ and $c_j$ (look up Finite Fourier Transform).
Thus your first example can be written as $$a_n = 4-\cos \left( \pi\,j/2 \right) -\cos \left( \pi\,j \right) -\cos \left( 3\,\pi\,j/2 \right) -\sin \left( \pi\,j/2 \right) +\sin \left( 3\,\pi\,j/2 \right) $$