7

While the function x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] can be obtained by integration of the integrand $\frac{1}{\sqrt{1-2x^3}}$ in Mathematica 12 as observed by Bob Hanlon, it cannot be obtained in Mathematica 11.2 and earlier versions, even though D[x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3],x] yields $\frac{1}{\sqrt{1-2x^3}}$ also in version 11.2....


6

A fix is to give System`Private`DerivativeX the NHoldAll attribute (which it probably should have, since it seems to be used as a dummy indexed variable): SetAttributes[System`Private`DerivativeX, NHoldAll] f[x_] = BesselI[0, 1.0 x]; f'[x] (* 1. BesselI[1, 1. x] *)


5

Here's a refactoring of the OP's code & Rolf's compile idea. Basically I tried to implement the idea in my comment: Also, BesselJ is not compilable, so it slows down the compiled function with call-backs to the main kernel. You could pass a list of Bessel function values as an argument to Compile. This would reduce the number of Bessel function calls ...


5

Using Compile speeds things up: λ = 0.500; k = (2 π)/λ; ρ[x_, y_] := Sqrt[x^2 + y^2]; ϕ[x_, y_] := ArcTan[x, y]; Ε2[n_, x_, y_, z_] := Exp[I k z + I n ϕ[x, y] + (I k ρ[x, y]^2)/(4 z)]* Sqrt[π/2]*(-I)^(Abs[n]/2)*Sqrt[(k*ρ[x, y]^2)/(4 z)]*(BesselJ[(Abs[n] - 1)/2, (k ρ[x, y]^2)/(4 z)] - I*BesselJ[(Abs[n] + 1)/2, (k ρ[x, y]^2)/(4 z)]); U = With[{expr = ((Exp[I*...


5

Try the replacement u=x^3. Then dx=du/3u^2/3and Integrate[1/3 u^(-2/3)/Sqrt[1 - 2 u], u] /. u -> x^3 yields (* (x^3)^(1/3) Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] *) Have fun!


5

Works well in version 12 $Version (* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *) int = Integrate[1/Sqrt[1 - 2 x^3], x] (* x Hypergeometric2F1[1/3, 1/2, 4/3, 2 x^3] *) The derivative returns the original expression D[int, x] (* 1/Sqrt[1 - 2 x^3] *)


4

The 5 Heun functions will be available in version 12.1 of Mathematica, so that we will be able to solve Heun-type of equations in build-in Mathematica functions.


3

The function being plotted is FullSimplify[DSolveValue[x y''[x] + (x + 1 - m) y'[x] - q y[x] == 0, y[x], x] /. {C[1] -> 0, C[2] -> Gamma[-q]/Gamma[m - q]}] E^-x x^m Hypergeometric1F1Regularized[1 + q, 1 + m, x] Surprisingly, another FunctionExpand[]/FullSimplify[] roundtrip gives a different, yet equivalent expression: FullSimplify[...


2

Unfortunately, no explanantion of the function PhysicalQ is given. What does it mean? If you had browsed the Wolfram Functions site more thoroughly, you would have found the "Primary Definition" page for ThreeJSymbol[], and then you would have seen the required definitions for PhysicalQ[] and the associated function TriangularQ[] (which yarchik has astutely ...


2

The result of cq = 1/6; m = 1/2; eq0 = x y''[x] + (x + 1 - m) y'[x] - q y[x]; so = AsymptoticDSolveValue[eq0 == 0, y[x], {x, Infinity, 7}]; so /. {C[1] -> 0, C[2] -> Gamma[-q]/Gamma[m - q]} *(1/Gamma[1/3])E^-x (-(1993326710375/(688747536 x^(23/3))) + 64899009175/(153055008 x^(20/3)) - 103178075/(1417176 x^(17/3)) + 2377375/(157464 x^(14/3)) - 8645/(...


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