Create algebraic function with finite power expansions?

In summary, it is easy to design a function with a finite power expansion using fractional polynomials. Just substitute w into f(z,w) 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
  • #1
aheight
321
109
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:
  • #3
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
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 to Create algebraic function with finite power expansions?

What is an algebraic function with finite power expansions?

An algebraic function with finite power expansions is a mathematical function that can be written as a finite sum of powers of its independent variable. This means that the function can be expressed using only addition, subtraction, multiplication, and division operations, and the powers of the independent variable are limited to whole numbers.

Can all algebraic functions be represented with finite power expansions?

No, not all algebraic functions can be represented with finite power expansions. Some functions, such as logarithms and trigonometric functions, require infinite power expansions to be fully represented. However, many common algebraic functions, like polynomials and rational functions, can be expressed with finite power expansions.

How do you create an algebraic function with finite power expansions?

To create an algebraic function with finite power expansions, you must first identify the pattern or rule that the function follows. This can often be done by looking at the given inputs and outputs of the function and finding a relationship between them. Once the pattern is identified, you can use it to write out the terms of the function, using the independent variable raised to different powers for each term.

What are some examples of algebraic functions with finite power expansions?

Some examples of algebraic functions with finite power expansions include linear functions (such as y = 2x + 3), quadratic functions (such as y = x^2 + 2x + 1), and rational functions (such as y = (x+1)/(x-2)). These functions can all be written as a finite sum of powers of the independent variable.

Why are finite power expansions useful in algebra?

Finite power expansions are useful in algebra because they allow us to easily manipulate and analyze complex functions. By breaking a function down into simpler terms, we can better understand its behavior and make predictions about its outputs for different inputs. Finite power expansions also make it easier to graph functions and find important points, such as zeros and extrema.

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