Can someone help me understand how a linear DE with rational coefficients in x is created for an algebraic function y(x) given implicitly by the equation:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]f(x,y)=a_0(z)+a_1(z)y+a_2(z)y^2+\cdots+a_n(x)y^n[/tex]

where a_i(z) are polynomials. This is described in the paper, "On Expansion of Algebraic Functions in Power and Puiseux Series I" by D.V and G.V Chudnovsky which I have a PDF copy but can't attach since it's 3 Mb and exceeds the 1.9 Mb limit but hopefully I can explain the part I'm having the problem with: It pertains to solving a diophantine equation in f(x,y) and f_y.

So we let:

[tex]P=f(x,y)[/tex]

[tex]Q=\frac{\partial f}{\partial y}[/tex]

and if we let:

[tex]r=\text{gcd}(P,Q)[/tex]

then according to the paper, we can solve a diophantine equation:

[tex]AP+BQ=r[/tex]

for A(x,y) and B(x,y) which are suppose to be polynomials in x and y. However, if I first solve for the greatest common denominator for f and f_y say for

[tex]f(x,y)=myFunction=1 + x - 2 x y + (3 - x) y^3[/tex]

using Mathematica's PolynomialGCD function, I get gcd=1. Ok, now when I try and solve for A and B using Mathematica's Solve:

Solve[a myFunction+b D[myFunction,y]==1,{a,b}]

I don't get polynomials but:

[tex]\left\{\left\{b\to -\frac{1}{2 x-9 y^2+3 x y^2}-\frac{a \left(1+x-2 x y+(3-x) y^3\right)}{-2 x+3 (3-x) y^2}\right\}\right\}[/tex]

which is just the diophantine equation re-expressed in terms of b. Can someone explain to me what I'm doing wrong? Also, if someone wants to look at the paper and can explain to me how to post it here, I'll do so.

Ok thanks guys,

Jack

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# Reducing algebraic function to DE

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