Skip to main content

complex - Why doesn't FullSimplify drop the Re function from an expression known to be real?


For some reason Mathematica does not properly simplify this expression:


In[7]:= FullSimplify[ArcTan[-Re[x + z], y], (x | y | z) \[Element] Reals]
Out[7]= ArcTan[-Re[x + z], y]


Obviously, if x and z are real, then so is x+z, so Re[x + z] should be replaced by x + z. Strangely enough, dropping any small part of the input fixes the problem, here are some examples.
No minus sign:


In[8]:= FullSimplify[ ArcTan[Re[x + z], y], (x | y | z) \[Element] Reals]
Out[8]= ArcTan[x + z, y]

No z:


In[9]:= FullSimplify[ArcTan[-Re[x], y], (x | y | z) \[Element] Reals]
Out[9]= ArcTan[-x, y]

No y:



In[10]:= FullSimplify[ArcTan[-Re[x + z]], (x | y | z) \[Element] Reals]
Out[10]= -ArcTan[x + z]

Of course I can just drop the Re function manually, but this is just a small fragment of the actual expression I'm trying to simplify, and I would like to avoid going though the whole expression looking for this specific pattern.
Anyone knows how to fix this? Is this a bug or what? (I'm using version 8.0.4.0)



Answer



The problem is due to Mathematica thinking that the version with the Re[] is actually simpler. This is because the default complexity function is more or less LeafCount[], and


In[332]:= ArcTan[-Re[x+z],y]//FullForm
Out[332]//FullForm= ArcTan[Times[-1,Re[Plus[x,z]]],y]


whereas


In[334]:= ArcTan[-x-z,y]//FullForm
Out[334]//FullForm= ArcTan[Plus[Times[-1,x],Times[-1,z]],y]

Here is a function that counts leaves without penalizing negation:


In[382]:= f3[e_]:=(LeafCount[e]-2Count[e,Times[-1,_],{0,Infinity}])
{LeafCount[x],LeafCount[-x],f3[x],f3[-x]}
Out[383]= {1,3,1,1}

If you tell mathematica to simplify using this complexity function then you get the expected result:



FullSimplify[ArcTan[-Re[x+z],y],(x|y|z)\[Element]Reals,ComplexityFunction->f3]


Out[375]= ArcTan[-x-z,y]



Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....