Skip to main content

Series expansion of expressions with Log and PolyLog functions (again)


I think that I discovered another (c.f. Series expansion of expressions with Log and PolyLog functions) issue related to Series and PolyLog.


Consider the following expression


exp = (-16*Pi^2*Log[1 + Sqrt[x]])/3 + (16*Pi^2*Log[1 + Sqrt[x]])/
x^4 + (16*Pi^2*Log[1 + Sqrt[x]])/(3*x^2) + (32*Pi^2*
Log[1 + Sqrt[x]])/(3*x) + (32*Pi^2*x*Log[1 + Sqrt[x]])/

3 - (16*Log[1 + Sqrt[x]]^3)/3 + (16*Log[1 + Sqrt[x]]^3)/
x^4 + (16*Log[1 + Sqrt[x]]^3)/(3*x^2) + (32*Log[1 + Sqrt[x]]^3)/(3*
x) + (32*x*Log[1 + Sqrt[x]]^3)/3 + (16*Pi^2*Log[1 - x])/
3 - (16*Pi^2*Log[1 - x])/
x^4 - (16*Pi^2*Log[1 - x])/(3*x^2) - (32*Pi^2*Log[1 - x])/(3*
x) - (32*Pi^2*x*Log[1 - x])/3 +
16*Log[1 + Sqrt[x]]^2*
Log[1 - x] - (48*Log[1 + Sqrt[x]]^2*Log[1 - x])/
x^4 - (16*Log[1 + Sqrt[x]]^2*Log[1 - x])/
x^2 - (32*Log[1 + Sqrt[x]]^2*Log[1 - x])/x -

32*x*Log[1 + Sqrt[x]]^2*Log[1 - x] -
16*Log[1 + Sqrt[x]]*
Log[1 - x]^2 + (48*Log[1 + Sqrt[x]]*Log[1 - x]^2)/
x^4 + (16*Log[1 + Sqrt[x]]*Log[1 - x]^2)/
x^2 + (32*Log[1 + Sqrt[x]]*Log[1 - x]^2)/x +
32*x*Log[1 + Sqrt[x]]*Log[1 - x]^2 + (16*Log[1 - x]^3)/
3 - (16*Log[1 - x]^3)/
x^4 - (16*Log[1 - x]^3)/(3*x^2) - (32*Log[1 - x]^3)/(3*x) - (32*x*
Log[1 - x]^3)/3 -
8*Log[1 + Sqrt[x]]^2*Log[x] + (24*Log[1 + Sqrt[x]]^2*Log[x])/

x^4 + (8*Log[1 + Sqrt[x]]^2*Log[x])/
x^2 + (16*Log[1 + Sqrt[x]]^2*Log[x])/x +
16*x*Log[1 + Sqrt[x]]^2*Log[x] +
16*Log[1 + Sqrt[x]]*Log[1 - x]*
Log[x] - (48*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/
x^4 - (16*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/
x^2 - (32*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x])/x -
32*x*Log[1 + Sqrt[x]]*Log[1 - x]*Log[x] -
8*Log[1 - x]^2*Log[x] + (24*Log[1 - x]^2*Log[x])/
x^4 + (8*Log[1 - x]^2*Log[x])/x^2 + (16*Log[1 - x]^2*Log[x])/x +

16*x*Log[1 - x]^2*Log[x] -
32*PolyLog[3, 1 - Sqrt[x]] + (96*PolyLog[3, 1 - Sqrt[x]])/
x^4 + (32*PolyLog[3, 1 - Sqrt[x]])/
x^2 + (64*PolyLog[3, 1 - Sqrt[x]])/x +
64*x*PolyLog[3, 1 - Sqrt[x]] -
32*PolyLog[3, Sqrt[x]] + (96*PolyLog[3, Sqrt[x]])/
x^4 + (32*PolyLog[3, Sqrt[x]])/x^2 + (64*PolyLog[3, Sqrt[x]])/x +
64*x*PolyLog[3, Sqrt[x]] -
32*PolyLog[
3, -(Sqrt[x]/(1 - Sqrt[x]))] + (96*

PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/
x^4 + (32*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/
x^2 + (64*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))])/x +
64*x*PolyLog[3, -(Sqrt[x]/(1 - Sqrt[x]))] +
32*Zeta[3] - (96*Zeta[3])/x^4 - (32*Zeta[3])/x^2 - (64*Zeta[3])/x -
64*x*Zeta[3]

exp can be made much shorter by applying Simplify to it, but in my case it is a part of a much larger expression and applying Simplify to the whole expression makes little sense.


In fact, for x between 0 and 1 exp vanishes identically. Therefore, if I expand this expression in x around 0 the expected result is zero. With Mathematica 8 and Mathematica 9 I get the desired result so that


Series[exp, {x, 0, 0}] // Normal


yields


0

However, starting with Mathematica 10 (I have version 10.3 here) something strange is going on. For some reason


Series[exp, {x, 0, 0}] // Normal

returns


-3201281/105840 + 2*Pi^2 + (84*(-3 + Log[x]))/(5*Sqrt[x]) + (1089*Log[x])/70


but


Series[Simplify[exp], {x, 0, 0}] // Normal

still gives me the correct result. The same behavior is observed with Mathematica 11.0 and 11.2. Using Assumptions -> x > 0 or Assumptions -> 0 < x < 1 does not change anything.


To me this very much looks like a bug. In any case I believe that the result returned by Series should not depend on whether one have applied Simplify to the initial expression or not. However, as this behavior might also be caused by a misunderstanding from my side, I'd like to hear the opinion of other users from this site first.




Comments

Popular posts from this blog

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-...

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...

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...