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Fungrim entry: 4d1365

(zk)=i=0ks ⁣(k,i)zik!{z \choose k} = \sum_{i=0}^{k} s\!\left(k, i\right) \frac{{z}^{i}}{k !}
Assumptions:zC  and  kZ0z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
{z \choose k} = \sum_{i=0}^{k} s\!\left(k, i\right) \frac{{z}^{i}}{k !}

z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
Binomial(nk){n \choose k} Binomial coefficient
Sumnf(n)\sum_{n} f(n) Sum
StirlingS1s ⁣(n,k)s\!\left(n, k\right) Signed Stirling number of the first kind
Powab{a}^{b} Power
Factorialn!n ! 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(Binomial(z, k), Sum(Mul(StirlingS1(k, i), Div(Pow(z, i), Factorial(k))), For(i, 0, k)))),
    Variables(z, k),
    Assumptions(And(Element(z, CC), Element(k, ZZGreaterEqual(0)))))

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2021-03-15 19:12:00.328586 UTC