From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: simplifying nested radicals
Date: 3 Jan 2000 23:44:29 GMT
Newsgroups: sci.math.symbolic

In article <84r6h3$poh$1@bambi.zdv.Uni-Mainz.DE>,
Alexander Frink <frink@BibServer.Physik.Uni-Mainz.DE> wrote:
>Could anybody please give me a hint how to show that the 
>following expression f with nested radicals is zero for y>0, x>y ?
>
>S:=(x+y+2)*(x+y-2)*(x-y-2)*(x-y+2);
>A:=1/2*((x^2+y^2-4)+sqrt(S));
>B:=1/2*((x^2-y^2)^2-4*(x^2+y^2)+(x^2-y^2)*sqrt(S));
>C:=1/2*((x^2-y^2)^2-4*(x^2+y^2)-(x^2-y^2)*sqrt(S));
>f:=(x+y)*(x*y-A)+(y^2-x^2+sqrt(B)-sqrt(C))*sqrt(A)+x*sqrt(B)+y*sqrt(C);
>
>Digits:=100;
>subs(x=3.8,y=1.4,f); # gives -0.16*10^(-97)

Define  k, l  to be the square roots of  (x+y-2)/(x+y+2)  and  (x-y-2)/(x-y+2)
respectively. (In your case,  k=2/3 and l=1/sqrt(11)  will do.)
Then we can do everything in terms of  k  and  l. Note that your expression
does NOT vanish unless the "right" square roots are taken; as you can
see I hand-edited  the formulas below to change the choices. These may
or may not be the right ones for you, depending on how your system selects
a symbolic square root. I mean the choices which make S2, A2, B2, C2
positive when  k=2/3, l=1/sqrt(11).

mysubs:=solve({k^2=(x+y-2)/(x+y+2), l^2=(x-y-2)/(x-y+2)}, {x,y}):
S:=(x+y+2)*(x+y-2)*(x-y-2)*(x-y+2);
S2:=simplify(factor(subs(mysubs,sqrt(S))), symbolic);
#use this instead of sqrt(S)
A:=1/2*((x^2+y^2-4)+S2);
A2:=-simplify(factor(subs(mysubs,sqrt(A))), symbolic); 
#use this instead of sqrt(A)
B:=1/2*((x^2-y^2)^2-4*(x^2+y^2)+(x^2-y^2)*S2);
B2:=simplify(factor(subs(mysubs,sqrt(B))), symbolic);
#use this instead of sqrt(B)
C:=1/2*((x^2-y^2)^2-4*(x^2+y^2)-(x^2-y^2)*S2);
C2:=-simplify(factor(subs(mysubs,sqrt(C))), symbolic); 
#use this instead of sqrt(C)
f:=(x+y)*(x*y-A)+(y^2-x^2+B2-C2)*A2+x*B2+y*C2;
factor(subs(mysubs,f));
                                       0


But this requires, for example, taking  A2  to be this square root:
    -2*(l+k)*(k*l-1)/(l-1)/(l+1)/(k-1)/(k+1)
Clearly for some choices of  l  and  k  this will be negative, and is
arguably the "wrong" choice. Perhaps this is why your symbolic packages
had some trouble.

dave