# Tag Info

5

The probability that the OP seeks is known as the multivariate Normal orthant probability. Correctly, for the $n=3$ cased posed here, the general integral DOES in fact have a closed -form solution, though Mma cannot (currently) obtain it. In particular, given a zero mean vector and variance-covariance matrix: \Sigma =\left( \begin{array}{ccc} 1 & ...

2

The general integral does not have a closed form solution, so use NIntegrate: mu = {0, 0, 0}; sigma = {{2, 1, 1}, {1, 2, 1}, {1, 1, 2}}; NIntegrate[ PDF[MultinormalDistribution[mu, sigma], {x, y, z}], {x, 0, ∞}, {y, 0, ∞}, {z, 0, ∞}] (* 0.25 *) Check: mu = {0, 0, 0}; sigma2 = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; NIntegrate[ ...

3

The two integrals are not equivalent, as can be seen from D[Tan[u], u] (* Sec[u]^2 *) The transformation from t to u takes the second integral from Integrate[1/Sqrt[1 - (nu t)^2], t] to Integrate[Sec[u]^2/Sqrt[1 - (nu Tan[u])^2], u]

1

a fix, specially treating the end interval: NBSpline[i_Integer, 0, knots_?(VectorQ[#, NumericQ] && OrderedQ[#] &), u_] /; i <= Length[knots] - 2 := ( If[ knots[[i + 2]] != knots[[-1]], Piecewise[{{1, knots[[i + 1]] <= u < knots[[i + 2]]}}, 0], Piecewise[{{1, knots[[i + 1]] <= u <= knots[[i + 2]]}}, 0] ]); ...

2

Perform the line integral over the straight-line path from the origin to $(x,y)$ via the parametrization $u(t)=(tx,ty)$ for $0\le t\le1$. Then $\mathrm du=(x,y)\,\mathrm dt$, so the integral can be computed as Integrate[b[t x, t y].{x, y}, {t, 0, 1}] Note that if $b = \nabla a$, from $b$ one can only recover $a$ up to a constant. In this case, since we ...

0

I actually went for this in the end: recursionNumber = 10; range = 100; f[list_, a_] := list - Take[Riffle[Table[0, {x, plist[[a]] range}], list/plist[[a]], plist[[a]]], range] plist = Prime[Range[range]]; list = DeleteCases[Fold[f, Table[1/n, {n, range}], Range[recursionNumber]], 0]; Plus @@ Join[{1}, ...

4

Hi Martin: Here is a partial solution using FoldList, along with a helper function fz defined as fz[z_, p_Integer] := Join[{1}, Cases[Apply[List,Expand[Total[z(1-p^-s)]]/.Power[m_,Times[e_,s]]:>(m^-e)^-s], _Integer^-s] ] I use fz to multiply the current expansion z by 1-p^-s, for prime p, then mess with the expression to get a ...

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