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Fungrim entry: 1eeccf

atan(x)k=02N+1(1)kx2k+12k+1\operatorname{atan}(x) \ge \sum_{k=0}^{2 N + 1} \frac{{\left(-1\right)}^{k} {x}^{2 k + 1}}{2 k + 1}
Assumptions:x[0,)  and  NZ0x \in \left[0, \infty\right) \;\mathbin{\operatorname{and}}\; N \in \mathbb{Z}_{\ge 0}
\operatorname{atan}(x) \ge \sum_{k=0}^{2 N + 1} \frac{{\left(-1\right)}^{k} {x}^{2 k + 1}}{2 k + 1}

x \in \left[0, \infty\right) \;\mathbin{\operatorname{and}}\; N \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
Atanatan(z)\operatorname{atan}(z) Inverse tangent
Sumnf(n)\sum_{n} f(n) Sum
Powab{a}^{b} Power
ClosedOpenInterval[a,b)\left[a, b\right) Closed-open interval
Infinity\infty Positive infinity
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(GreaterEqual(Atan(x), Sum(Div(Mul(Pow(-1, k), Pow(x, Add(Mul(2, k), 1))), Add(Mul(2, k), 1)), For(k, 0, Add(Mul(2, N), 1))))),
    Variables(x, N),
    Assumptions(And(Element(x, ClosedOpenInterval(0, Infinity)), Element(N, ZZGreaterEqual(0)))))

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2021-03-15 19:12:00.328586 UTC