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Fungrim entry: da0f15

Wk ⁣(z)(L1L2+n=0N1m=1M1(1)nm![n+mn+1]σnτm)4τ(4σ)N+(4τ)M(14σ)(14τ)   where L1=log(z)+2πik,L2=log ⁣(L1),σ=1L1,τ=L2L1\left|W_{k}\!\left(z\right) - \left({L}_{1} - {L}_{2} + \sum_{n=0}^{N - 1} \sum_{m=1}^{M - 1} \frac{{\left(-1\right)}^{n}}{m !} \left[{n + m \atop n + 1}\right] {\sigma}^{n} {\tau}^{m}\right)\right| \le \frac{4 \left|\tau\right| {\left(4 \left|\sigma\right|\right)}^{N} + {\left(4 \left|\tau\right|\right)}^{M}}{\left(1 - 4 \left|\sigma\right|\right) \left(1 - 4 \left|\tau\right|\right)}\; \text{ where } {L}_{1} = \log(z) + 2 \pi i k,\,{L}_{2} = \log\!\left({L}_{1}\right),\,\sigma = \frac{1}{{L}_{1}},\,\tau = \frac{{L}_{2}}{{L}_{1}}
Assumptions:NZ0andMZ0andkZandzC{0}andσ<14andτ<14and(k0orz>1)   where L1=log(z)+2πik,L2=log ⁣(L1),σ=1L1,τ=L2L1N \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, M \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z} \,\mathbin{\operatorname{and}}\, z \in \mathbb{C} \setminus \left\{0\right\} \,\mathbin{\operatorname{and}}\, \left|\sigma\right| < \frac{1}{4} \,\mathbin{\operatorname{and}}\, \left|\tau\right| < \frac{1}{4} \,\mathbin{\operatorname{and}}\, \left(k \ne 0 \,\mathbin{\operatorname{or}}\, \left|z\right| > 1\right)\; \text{ where } {L}_{1} = \log(z) + 2 \pi i k,\,{L}_{2} = \log\!\left({L}_{1}\right),\,\sigma = \frac{1}{{L}_{1}},\,\tau = \frac{{L}_{2}}{{L}_{1}}
TeX:
\left|W_{k}\!\left(z\right) - \left({L}_{1} - {L}_{2} + \sum_{n=0}^{N - 1} \sum_{m=1}^{M - 1} \frac{{\left(-1\right)}^{n}}{m !} \left[{n + m \atop n + 1}\right] {\sigma}^{n} {\tau}^{m}\right)\right| \le \frac{4 \left|\tau\right| {\left(4 \left|\sigma\right|\right)}^{N} + {\left(4 \left|\tau\right|\right)}^{M}}{\left(1 - 4 \left|\sigma\right|\right) \left(1 - 4 \left|\tau\right|\right)}\; \text{ where } {L}_{1} = \log(z) + 2 \pi i k,\,{L}_{2} = \log\!\left({L}_{1}\right),\,\sigma = \frac{1}{{L}_{1}},\,\tau = \frac{{L}_{2}}{{L}_{1}}

N \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, M \in \mathbb{Z}_{\ge 0} \,\mathbin{\operatorname{and}}\, k \in \mathbb{Z} \,\mathbin{\operatorname{and}}\, z \in \mathbb{C} \setminus \left\{0\right\} \,\mathbin{\operatorname{and}}\, \left|\sigma\right| < \frac{1}{4} \,\mathbin{\operatorname{and}}\, \left|\tau\right| < \frac{1}{4} \,\mathbin{\operatorname{and}}\, \left(k \ne 0 \,\mathbin{\operatorname{or}}\, \left|z\right| > 1\right)\; \text{ where } {L}_{1} = \log(z) + 2 \pi i k,\,{L}_{2} = \log\!\left({L}_{1}\right),\,\sigma = \frac{1}{{L}_{1}},\,\tau = \frac{{L}_{2}}{{L}_{1}}
Definitions:
Fungrim symbol Notation Short description
Absz\left|z\right| Absolute value
LambertWWk ⁣(z)W_{k}\!\left(z\right) Lambert W-function
Sumnf(n)\sum_{n} f(n) Sum
Powab{a}^{b} Power
Factorialn!n ! Factorial
StirlingCycle[nk]\left[{n \atop k}\right] Unsigned Stirling number of the first kind
Loglog(z)\log(z) Natural logarithm
ConstPiπ\pi The constant pi (3.14...)
ConstIii Imaginary unit
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
ZZZ\mathbb{Z} Integers
CCC\mathbb{C} Complex numbers
Source code for this entry:
Entry(ID("da0f15"),
    Formula(Where(LessEqual(Abs(Sub(LambertW(k, z), Parentheses(Add(Sub(Subscript(L, 1), Subscript(L, 2)), Sum(Sum(Mul(Mul(Mul(Div(Pow(-1, n), Factorial(m)), StirlingCycle(Add(n, m), Add(n, 1))), Pow(sigma, n)), Pow(tau, m)), For(m, 1, Sub(M, 1))), For(n, 0, Sub(N, 1))))))), Div(Add(Mul(Mul(4, Abs(tau)), Pow(Mul(4, Abs(sigma)), N)), Pow(Mul(4, Abs(tau)), M)), Mul(Sub(1, Mul(4, Abs(sigma))), Sub(1, Mul(4, Abs(tau)))))), Equal(Subscript(L, 1), Add(Log(z), Mul(Mul(Mul(2, ConstPi), ConstI), k))), Equal(Subscript(L, 2), Log(Subscript(L, 1))), Equal(sigma, Div(1, Subscript(L, 1))), Equal(tau, Div(Subscript(L, 2), Subscript(L, 1))))),
    Variables(k, z, N, M),
    Assumptions(Where(And(Element(N, ZZGreaterEqual(0)), Element(M, ZZGreaterEqual(0)), Element(k, ZZ), Element(z, SetMinus(CC, Set(0))), Less(Abs(sigma), Div(1, 4)), Less(Abs(tau), Div(1, 4)), Or(Unequal(k, 0), Greater(Abs(z), 1))), Equal(Subscript(L, 1), Add(Log(z), Mul(Mul(Mul(2, ConstPi), ConstI), k))), Equal(Subscript(L, 2), Log(Subscript(L, 1))), Equal(sigma, Div(1, Subscript(L, 1))), Equal(tau, Div(Subscript(L, 2), Subscript(L, 1))))))

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2019-10-05 13:11:19.856591 UTC