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Fungrim entry: 02ee06

(z)k+m=(z)k(z+k)m\left(z\right)_{k + m} = \left(z\right)_{k} \left(z + k\right)_{m}
Assumptions:zCandkZ0andmZ0z \in \mathbb{C} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, m \in \mathbb{Z}_{\ge 0}
\left(z\right)_{k + m} = \left(z\right)_{k} \left(z + k\right)_{m}

z \in \mathbb{C} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, m \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
RisingFactorial(z)k\left(z\right)_{k} Rising factorial
CCC\mathbb{C} Complex numbers
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(Equal(RisingFactorial(z, Add(k, m)), Mul(RisingFactorial(z, k), RisingFactorial(Add(z, k), m)))),
    Variables(z, k, m),
    Assumptions(And(Element(z, CC), Element(k, ZZGreaterEqual(0)), Element(m, ZZGreaterEqual(0)))))

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Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-10-05 13:11:19.856591 UTC