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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

You can do Integrate[(1 - x^-y)/(1 - x), {x, 0, 1}] (* HarmonicNumber[-y], Re[y] < 1*)] and then (HarmonicNumber[-y] /. y -> 1/5) // FunctionExpand // FullSimplify (* 1/2 Sqrt[1 + 2/Sqrt[5]] \[Pi] - 1/2 Sqrt[5] ArcCoth[Sqrt[5]] - (5 Log[5])/4*)


4

This is a major bug in version 12.0.0. Frankly, I am puzzled by why they didn't release a 12.0.1 to fix this ... for me it is a recurring major inconvenience. The file in which the settings are stored can be found in this directory: SystemOpen@FileNameJoin@{$UserBaseDirectory, "FrontEnd"} Close Mathematica before editing it. First try init.m. If it ...


2

Since it was too long for a comment, I transcribe everything here together: f1[x_] := Sqrt[(5 + Sqrt[5])/2] ArcTan[(1 - Sqrt[5] + 4 x^(1/5))/Sqrt[2 (5 + Sqrt[5])]] f2[x_] := Sqrt[(5 - Sqrt[5])/2] ArcTan[(1 + Sqrt[5] + 4 x^(1/5))/Sqrt[2 (5 - Sqrt[5])]] f3[x_] := -(5 - Sqrt[5])/4 Log[2 + (1 - Sqrt[5]) x^(1/5) + 2 x^(2/5)] f4[x_] := -(5 + Sqrt[5])/4 Log[2 + (1 ...


2

Not certain what is going on but you can work around with Interpolation. ifoo = Interpolation[ha, InterpolationOrder -> 1]; Show[ ListPointPlot3D[ha, PlotRange -> All, AxesLabel -> {"x", "y", "z"}, ImageSize -> Large], Plot3D[ifoo[x, y], Evaluate[Sequence @@ MapThread[Prepend, {ifoo["Domain"], {x, y}}]], PlotStyle -> Opacity[...


1

It might be interesting how we can simplify the expression obtained by Mathematica with the method 2) of my analysis: iy1 = -Log[5] - (-1)^(1/5) Log[1 + (-1)^(1/5)] + (-1)^(2/5) Log[1 - (-1)^(2/5)] - (-1)^(3/5) Log[1 + (-1)^(3/5)] + (-1)^(4/5) Log[1 - (-1)^(4/5)] The basic action is a replacement exemplified here Log[1 + (-1)^(1/5)] /. Log[a_] -> ...


1

There is a simple workaround: ClearAll[a, x]; f = (4 x^2)/((-a^2 - 4 x^2 + a Sqrt[a^2 + 4 x^2]) (a^2 + 4 x^2 + a Sqrt[a^2 + 4 x^2])); Series[f, {x, 0, 1}, Assumptions -> Re[a] > 0] *-(1/a^2)+O[x]^2 * Series[f, {x, 0, 1}, Assumptions -> Re[a] < 0] *-(1/a^2)+O[x]^2 * The case $\Re a=0$ makes a trouble.


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