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Fungrim entry: 71a264

sin2n+1 ⁣(z)=14nk=0n(1)n+k(2n+1k)sin ⁣((2n2k+1)z)\sin^{2 n + 1}\!\left(z\right) = \frac{1}{{4}^{n}} \sum_{k=0}^{n} {\left(-1\right)}^{n + k} {2 n + 1 \choose k} \sin\!\left(\left(2 n - 2 k + 1\right) z\right)
Assumptions:zC  and  nZ0z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; n \in \mathbb{Z}_{\ge 0}
\sin^{2 n + 1}\!\left(z\right) = \frac{1}{{4}^{n}} \sum_{k=0}^{n} {\left(-1\right)}^{n + k} {2 n + 1 \choose k} \sin\!\left(\left(2 n - 2 k + 1\right) z\right)

z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; n \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
Powab{a}^{b} Power
Sinsin(z)\sin(z) Sine
Sumnf(n)\sum_{n} f(n) Sum
Binomial(nk){n \choose k} Binomial coefficient
CCC\mathbb{C} Complex numbers
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(Equal(Pow(Sin(z), Add(Mul(2, n), 1)), Mul(Div(1, Pow(4, n)), Sum(Mul(Mul(Pow(-1, Add(n, k)), Binomial(Add(Mul(2, n), 1), k)), Sin(Mul(Add(Sub(Mul(2, n), Mul(2, k)), 1), z))), For(k, 0, n))))),
    Variables(z, n),
    Assumptions(And(Element(z, CC), Element(n, ZZGreaterEqual(0)))))

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2021-03-15 19:12:00.328586 UTC