Zetas and positive even integers
By using quadratic equation:
𝐴 𝑛
−1±√3𝑖 2
𝑥 =
−1±√3𝑖 2𝐴
𝑛 𝐴
𝑥 =
𝑛 𝑜𝑟 𝑥 =
(𝟕)
Where 𝑛 ∈ℤ + and A is a constant.
If some equation has all the roots shown in (𝟕) only and has another way of expressing the infinite polynomial, the mystery of the zetas of positive odd integers might be solved.
Conclusion
Generating function derived from the manipulation of sin(𝑥) can be used to calculate the zetas of positive even integers. However, its computation complexity means this is not a suitable method for computing large inputs. But it has potential to be used to derive some closed form for the unknown zetas of positive odd integers.
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