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6

Total[funca[a,#] & /@ #] & /@ {x,y} There are two Function expressions here which I will refer to as inner and outer. The inner function: funca[a,#] & Is Mapped to the sole argument of the outer function. It will transform a list or other expression like this: funca[a,#] & /@ foo[1, 2, 3] foo[funca[a,1], funca[a,2], funca[a,3]] ...


3

The approach that kguler suggest in your precedent question is completely suitable for the current one: FindSequenceFunction[Table[With[{ d1 = HypoexponentialDistribution[Flatten[Table[{λ, μ}, {i - 1}]]], d2 = HypoexponentialDistribution[{λ, μ}], d3 = ExponentialDistribution[μ], d = ExponentialDistribution[λ]}, ...


3

I think I might need more examples to figure out how to automate the level accurately. This works on the given example (in V10): Map[Inactivate, (1 + (1 + x^2)^2)^2, {2}] // Expand // Activate (* 1 + 2 (1 + x^2)^2 + (1 + x^2)^4 *) Why level 2? Let's look at the tree form of the expression: (1 + (1 + x^2)^2)^2 // TreeForm What we see is that the ...


3

You can use an equation as an assumption Simplify[λ (λ - 1), λ (λ - 1) == 0] 0


3

These are the conditions necessary for the expression to be Positive Assuming[0 < q < 1 && 0 < y < x < 1, FullSimplify@ Reduce[Positive[ 1/2 - (2 q (2 q + x^2 - y^2))/(2 q (q - 2) - x^2 + y^2)^2]]] Sqrt[2] + q == 2 || [...] and other more complicated solutions. If you need to know if that's always true then Resolve[ ...


2

You are invoking ForAll with vacuous conditions. Compare: ForAll[{x}, x == x + 1, Element[Abs[x], Reals]] (* output: True *) I think that is really all that's going on here. There is no x in the ForAll for which x==Infinity actually returns true, so it spits out true because the "all" in "for all" is the empty set. It's vacuous. Likewise in the limit, ...


2

Let's see how far we can get in expressing the double integral in terms of closed form expressions, hereby pursuing the path I indicated in my comment. The Integrand is f = ((1 - x - y + x y + x Log[x] - x y Log[x] + y Log[y] - x y Log[y] + x y Log[x] Log[y]) Log[1 + x y])/((1 - x) x (1 - y) y Log[x] Log[y]) and the integral in question is fi = ...


2

denom = 25 + 2 g + x^2; x0 = I/2 Sqrt[100 + 8 g]; Rather than substituting with x -> x0, take the limit result[g_] = Limit[(x - x0)/denom, x -> x0] -(I/(2*Sqrt[25 + 2*g])) result[1000] -(I/90) result[1] -(I/(6*Sqrt[3])) Apart works with symbolic expressions: (x - x0)/denom ((-(1/2))ISqrt[100 + 8*g] + x)/ (25 + 2*g + x^2) ...


1

Seeing as you're trying to evaluate hydrogen wavefunctions, note that the necessary special functions are already built-in, so you can skip the step of defining the special functions entirely, and just do this: ψ[n_, l_, m_, ρ_, θ_, ϕ_] := Sqrt[(2/(n a0))^3 (n - l - 1)!/(2 n (n + l)!)] Exp[-ρ/2] ρ^ l LaguerreL[n - l - 1, 2 l + 1, ρ] ...


1

This is not an answer to your question (hence the community tag) since I do not know why Integrate does not solve this, but to point out that the command Int solves this instantly with no problem. This is using Albert Rich Rubi package: ShowSteps = False; Int[(1 + (1 + 1/(2*Sqrt[x]))/(2*Sqrt[Sqrt[x] + x]))/(2*Sqrt[x + Sqrt[Sqrt[x] + x]]), x]


1

f[x_, y_, c_] /; x > 0 && y > 0 := (#.{Sqrt@x, Sqrt@y} < Sqrt[c]) & /@ Tuples[{-1, 1}, 2] plot[a_, b_, c_] := RegionPlot[Xor @@ f[x - a, y - b, c], {x, 0, 4 a}, {y, 0, 4 b}, BoundaryStyle -> {Thick, Blue}, PlotStyle -> Transparent, PlotRange -> {{0, 4 a .9}, {0, 4 ...


1

You could rationalize the equation. With[{conjugates = pm Sqrt[x - a] + pm2 Sqrt[y - b] == Sqrt[c] /. {{pm -> 1}, {pm -> -1}} /. {{pm2 -> 1}, {pm2 -> -1}} /. Equal -> Subtract}, eqn = 0 == Times @@ Flatten[conjugates] // Expand // PowerExpand ]; Block[{a = 2, b = 3, c = 4}, Print[eqn]; ContourPlot[Evaluate@eqn, {x, 0, 4 a}, {y, 0, ...



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