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Fungrim entry: c5dd9b

Pn ⁣(z)=12nk=0n(nk)2(z1)nk(z+1)kP_{n}\!\left(z\right) = \frac{1}{{2}^{n}} \sum_{k=0}^{n} {{n \choose k}}^{2} {\left(z - 1\right)}^{n - k} {\left(z + 1\right)}^{k}
Assumptions:nZ0  and  zCn \in \mathbb{Z}_{\ge 0} \;\mathbin{\operatorname{and}}\; z \in \mathbb{C}
P_{n}\!\left(z\right) = \frac{1}{{2}^{n}} \sum_{k=0}^{n} {{n \choose k}}^{2} {\left(z - 1\right)}^{n - k} {\left(z + 1\right)}^{k}

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

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2021-03-15 19:12:00.328586 UTC