Create algebraic function with finite power expansions?

  • I
  • Thread starter aheight
  • Start date
  • #1
315
107
How would I design a non-trivial algebraic function of degree 4 containing a branch at the origin with the (finite) power expansion:

##w(z)=1+0.5 z-1/4 z^{1/2}+3/4 z^{1/4}##?

having the form

## f(z,w)=a_0+a_1 w+a_2 w^2+a_3 w^3+a_4 w^4=0##

with the ##a_i ## ( preferably not fractional) polynomials? And if that is possible, can I design any function ##w(z)## with a finite power expansion with ##f## of any degree?

I don't know.
 
Last edited:

Answers and Replies

  • #2
18,423
8,260
Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
 
  • #3
315
107
I figured out it's easy to design one using fractional polynomials. Just substitute ##w(z)## into ##f(z,w)=a_0+w+w^2+w^3+w^4=0## for example and solve for ##a_0## to get:

##
w^4+w^3+w^2+w-\frac{1}{128} 237 z^{3/2}+\frac{z^{5/2}}{8}+\frac{z^{7/2}}{8}+\frac{105 z^{3/4}}{64}-\frac{249 z^{5/4}}{32}+\frac{129 z^{7/4}}{64}-\frac{99 z^{9/4}}{32}+\frac{9 z^{11/4}}{16}-\frac{3 z^{13/4}}{8}-\frac{z^4}{16}-\frac{23 z^3}{32}-\frac{545 z^2}{256}-\frac{15 \sqrt[4]{z}}{2}-\frac{981 z}{256}-\frac{25 \sqrt{z}}{8}-4=0
##

but I think that introduces 12 more cycles into the function: for each root of ##z^{1/4}## we have four values of ##w## or maybe we just have a reducible function ##w## made up of four factors each of degree four. Not sure.
 
Last edited:
  • #4
315
107
Here is a solution in Mathematica (see https://www.physicsforums.com/threa...ary-constants-with-solve.899472/#post-5660255 for some background) .

For example, given ##\text{myw}=z^{3/4}+z^{5/4}+\sqrt{z}+\sqrt[4]{z}+z+1##, the code produces:

##15 w^6+w^5 (-60 z-55)+w^4 \left(30 z^2+70 z+71\right)+w^3 \left(-50 z^2-34 z-34\right)+w^2 \left(-15 z^5-15 z^4-15 z^3-3 z^2+z+1\right)+w \left(-5 z^5-5 z^4-5 z^3+z^2+z+1\right)-z^5-z^4-z^3+z^2+z+1##

What remains is to prove the following hypothesis:

Given any fractional polynomial ##u(z)##, there exists a bivariate polynomial ##f(z,w)## (with integer powers of z and w) such that ##f(z,u)=0##. I believe this statement is true. Would anyone here suggest a means of proving this?

Thanks!

Code:
myw = 1 + z^(1/4) + z^(1/2) + z^(3/4) + z + z^(5/4)
fExponents = Exponent[myw, z, List]
fdegree = Length[fExponents];
matrixFlag = False;
While[! matrixFlag && fdegree < 10,
  {
 
   pdegree = fdegree;
   Quiet[Table[
     Subscript[s, n] =., {k, 0, fdegree}, {n, 20 k, 20 k + pdegree}]];
 
   theCoefficients =
    Table[Subscript[s, n], {k, 0, fdegree}, {n, 20 k, 20 k + pdegree}];
   thePolyTerms =
    Table[theCoefficients[[i, j]] z^(j - 1), {i, 1, fdegree + 1}, {j,
      1, pdegree + 1}];
   thePolys = (Plus @@ # &) /@ thePolyTerms;
   theFunctionForm =
    Plus @@ Flatten[
      Table[thePolys[[i]] w^(i - 1), {i, 1, fdegree + 1}]];
 
   expandedForm = Expand[theFunctionForm /. w -> myw];
   elist = Exponent[expandedForm, z, List];
   theEquations =
    Table[Coefficient[expandedForm, z, elist[[i]]], {i, 1,
      Length[elist]}];
   theSolution =
    Solve[Table[
      Coefficient[expandedForm, z, elist[[i]]] == 0, {i, 1,
       Length[elist]}], Flatten[theCoefficients], Method -> "Reduce"];
   val1 = theSolution // Flatten;
   val2 = Association[val1];
   clist = Flatten[theCoefficients];
   iList = Intersection[Flatten[theCoefficients], Keys[val2]];
   theNormalForm =
    Normal[CoefficientArrays[#, Flatten[theCoefficients]]] & /@
     theEquations;
   theCMatrix = #[[2]] & /@ theNormalForm;
   theMRank = MatrixRank[theCMatrix];
   Print["number of equations", Length[theEquations]];
   Print["number of variables: ", Length[Flatten[theCoefficients]]];
   Print["Matrix Rank: ", MatrixRank[theCMatrix]];
   If[theMRank < Length[Flatten[theCoefficients]],
    {
     Print[
      "Rank less than number of variables.  Computing function . . . \
"];
     theFreeVals = Complement[clist, iList];
     Print["free vars: ", theFreeVals];
     For[i = 1, i <= Length[theFreeVals], i++,
      Subscript[theFreeVals[[i, 1]], theFreeVals[[i, 2]]] = 1;
      ];
     Print["Function: ",
      polyForm[(theFunctionForm /. theSolution)[[1]], w]];
     matrixFlag = True;
     }
    ,
    fdegree++;
    ];
   }
  ];
 
Last edited:

Related Threads on Create algebraic function with finite power expansions?

Replies
2
Views
663
  • Last Post
Replies
1
Views
725
Replies
1
Views
2K
Replies
5
Views
5K
Replies
2
Views
2K
Replies
15
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
7
Views
2K
Top