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

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...