Application to linear independence

Theorem 1. The functions ( R_F(x,y,z) ), ( R_D(x,y,z) ), ( R_J(x,y,z,p) ), and ( (xyz)^-1/2 ) are linearly independent with respect to coefficients that are rational functions of ( x,y,z ), and p.

Proof. Let ( , , ), and ( ) be rational functions of ( x,y,z ), and ( p ). We need to prove that

iff ( , , ), and ( ) are identically 0. We may assume that these coefficients are polynomials since we can multiply all terms by the denominator of any rational function. As (p 0 ), () shows that (R_J(x,y,z,p) ) involves ( p ) while all other quantities are polynomials in p, whence ( 0 ). As ( z ) we have [ = az^m(1+O(1/z)), = bz^n(1+O(1/z)), ] where ( m ) and ( n ) are nonnegative integers and ( a ) and ( b ) are polynomials in x,y, and p. Using () and () and multiplying all terms by (2z^3/2 ), we find [ az^m+1 [ 8za+g + O ( zz ) ] + 3bz^n [ 8za+g -2 + O ( zz ) ] + 2 , (xy)^-1/2z 0. ] Cancellation of the leading terms in ( z ) requires ( az^m+1 + 3bz^n 0 ), implying ( n = m+1 ) and ( a -3b ) and leaving [ O(z^m z) - 6bz^m+1 + 2 , (xy)^-1/2z 0. ] Because the second term is of different order from the first and does not have a square root in common with the third, it follows that ( b ), whence also ( a 0 ). Since the leading terms of the polynomials ( ) and ( ) are identically 0, so too are ( ) and ( ). Finally, with only one term remaining in (), we have ( 0 ). ( )

It is an open question whether Theorem 1 is still true if the coefficients are algebraic functions instead of rational functions. However, polynomial coefficients suffice (see the first paragraph of [7, § 9.2,]) to prove that every elliptic integral can be expressed in terms of ( R_F, R_D, R_J ), and elementary functions.

Appendix