Skip to main content

simplifying expressions - Different outputs for Simplify


In at least one instance, the same Simplify command (without using options) produces very different results on two of my computers, one Windows 7 and the other Windows 10. Here is the code:


sf = Simplify;

r1 = (-1 + (-11 - a^2 + 2 a^4) b^2 + (8 - 3 a^2) b^3 + (-2 + a^2) b^4 - 
     b (-6 - 3 a^2 + a Sqrt[b (2 + (-2 + a^2) b)] + 
        2 a^3 Sqrt[b (2 + (-2 + a^2) b)]) - 
     a (Sqrt[b (2 + (-2 + a^2) b)] - 3 Sqrt[b^5 (2 + (-2 + a^2) b)] + Sqrt[

        b^7 (2 + (-2 + a^2) b)]))/(-1 + b - a^2 b + 
     a Sqrt[b (2 + (-2 + a^2) b)])^2;

This is what I got on one of the computers, the Windows 10 one:


Dr1 = D[r1, b] // sf 

(* (6 + 3 a^2 + 2 (-11 - a^2 + 2 a^4) b + 3 (8 - 3 a^2) b^2 + 
    4 (-2 + a^2) b^3 - (a (1 + (-2 + a^2) b))/Sqrt[b (2 + (-2 + a^2) b)] - (
    a b (1 + (-2 + a^2) b))/Sqrt[b (2 + (-2 + a^2) b)] - (
    2 a^3 b (1 + (-2 + a^2) b))/Sqrt[b (2 + (-2 + a^2) b)] - 

    a Sqrt[b (2 + (-2 + a^2) b)] - 2 a^3 Sqrt[b (2 + (-2 + a^2) b)] + (
    a b^6 (-7 - 4 (-2 + a^2) b))/Sqrt[b^7 (2 + (-2 + a^2) b)] + (
    3 a b^4 (5 + 3 (-2 + a^2) b))/Sqrt[b^5 (2 + (-2 + a^2) b)])/(-1 + b - 
    a^2 b + a Sqrt[
     b (2 + (-2 + a^2) b)])^2 - (2 (1 - a^2 + (a - 2 a b + a^3 b)/Sqrt[
      b (2 + (-2 + a^2) b)]) (-1 + (-11 - a^2 + 2 a^4) b^2 + (8 - 
         3 a^2) b^3 + (-2 + a^2) b^4 - 
      b (-6 - 3 a^2 + a Sqrt[b (2 + (-2 + a^2) b)] + 
         2 a^3 Sqrt[b (2 + (-2 + a^2) b)]) - 
      a (Sqrt[b (2 + (-2 + a^2) b)] - 3 Sqrt[b^5 (2 + (-2 + a^2) b)] + Sqrt[

         b^7 (2 + (-2 + a^2) b)])))/(-1 + b - a^2 b + 
    a Sqrt[b (2 + (-2 + a^2) b)])^3 *)

And this, much better expression is what I got on the other computer, the Windows 7 one:


D[r1, b] // sf

(* (2 (-1 + b)^2 b (4 (-2 + 9 a^2 - 8 a^4 + 2 a^6) b^4 - 
     3 a Sqrt[b (2 + (-2 + a^2) b)] - 
     b (-8 - 11 a^2 + 10 a Sqrt[b (2 + (-2 + a^2) b)] + 
        10 a^3 Sqrt[b (2 + (-2 + a^2) b)]) - 

     b^2 (24 - 14 a^2 - 14 a^4 - 29 a Sqrt[b (2 + (-2 + a^2) b)] + 
        14 a^3 Sqrt[b (2 + (-2 + a^2) b)] + 
        4 a^5 Sqrt[b (2 + (-2 + a^2) b)]) - 
     b^3 (-24 + 61 a^2 - 18 a^4 - 4 a^6 + 16 a Sqrt[b (2 + (-2 + a^2) b)] - 
        24 a^3 Sqrt[b (2 + (-2 + a^2) b)] + 
        8 a^5 Sqrt[b (2 + (-2 + a^2) b)])))/(Sqrt[
   b (2 + (-2 + a^2) b)] (-a b + Sqrt[b (2 + (-2 + a^2) b)])^3 (-1 + b - 
     a^2 b + a Sqrt[b (2 + (-2 + a^2) b)])^2) *)

Can I find the reason for this and make it the same on both computers? If so, how to do this? I may have changed global preferences for Simplify on the Windows 7 computer, but don't remember if or how or what I did concerning that. This behavior occurs both in Mathematica 11.2 and Mathematica 11.1.1.





Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more