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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] ...


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) ...


0

One method of obtaining a solution accurate to first order in δ, based on the comment by @Daniel Lichtblau is as follows: First, apply the function Normal[Series[# /. {gw -> gw0 + δ gw1, gy -> gy0 + δ gy1}, {δ, 0, 1}]] & to each equation to obtain them accurate to first order in δ. Second, set δ to zero in the resulting step #1 equations and Solve ...


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]}, ...


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]


0

I can't comment on whether this functionality is by design or by omission; however, to fix this behavior, one can use UnitConvert Quantity["Moles" / "Liters"] // UnitSimplify UnitConvert[%,"Molar"] (* 1 mol/L *) (* 1 M *)


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, ...


0

First of all, the underscore is not allowed in a variable name. Your expression can be written in Mathematica for instance format like this FullSimplify[((p1 Cp)/R) (V2 - V1) + ((V2 CV)/R) (p2 - p1), Assumptions -> {p1 V1^(gamma) == p2 V2^(gamma), gamma == Cp/CV}] (* Out[70]= (CV (-p1 + p2) V2 + Cp p1 (-V1 + V2))/R *) Next, even though it sounds ...


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]] ...


0

I would suggest you apply the FullSimplify on the levels starting from the lowest level. even if you have to apply the FullSimplify on some easy expressions but that will not take much time. try this and see if it help you: Replace[expression, x_ :> FullSimplify[x], {-2, -1}] you can change the levels range according to your need. if you can writ ...


0

An attempt at automating Kuba's (now deleted) answer, assuming that what is required is to map the function over the arguments of Plus and Power: MapAt[f, expression, Position[expression, Plus | Power] /. {{x__, 0} :> {x}, {0} -> {1}}] (Using a generic function f for demonstration purposes rather than FullSimplify.)


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 = ...


0

expr = ((x^2 + 1)^2 + 1)^2; Expand[Map[Defer, expr, {2}]] Expand[Map[HoldForm, expr, {2}]] Block[{df}, Expand[Map[df, expr, {2}]] /. df -> Identity] (* 1 + 2 (1 + x^2)^2 + (1 + x^2)^4 *) You can also use the second argument of Expand: Expand[expr, Cases[expr, _Power, {2}][[1]]] Expand[expr, _[___, _^_, ___]^_] (* 1 + 2 (1 + x^2)^2 + (1 + x^2)^4 *)


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 ...


0

There you go: CExpand[Power[exp_Plus, n_]] := Sum[Binomial[n, k] First@exp^(n - k) CExpand@Power[Drop[exp, 1], k], {k, 0, n}] CExpand[exp_] := exp Another idea, which works more generally, is to use Replace[exp,{a_Plus->CPlus@@a,b_Power->CPower@@b},{2,Infinity}] This replaces the build-in functions by some undefined ones, which Mathematica will ...


0

One more try: relations = SolveAlways[(b1 x + b2 x)/(b3 + (b4/b5) x) == (c1 x)/(c2 + c3 x), {x}][[4]] {b3 -> ((b1 + b2) c2)/c1, b4 -> ((b1 + b2) b5 c3)/c1} This gives nontrivial relations between parameters you should always keep for any x. Convert them to equations and remove some freedom restrictions = Equal @@@ relations; myChoice = ...


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[ ...



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