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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