The lack of handling root functions in hp prime

[yangyongkang]

This equation group, XCAS calculations fall into an infinite loop.

Code:

 f(x,y):=4*sqrt(16*y^2*tan(x/2)^8*(1/(tan(x/2)^2+1))^8+16*y^2*tan(x/2)^6*(1/(tan(x/2)^2+1))^8+16*y^2*tan(x/2)^7*(1/(tan(x/2)^2+1))^7+48*y^2*tan(x/2)^5*(1/(tan(x/2)^2+1))^7+4*y^2*tan(x/2)^6*(1/(tan(x/2)^2+1))^6-32*y^2*tan(x/2)^7*(1/(tan(x/2)^2+1))^7/(tan(x/2)^2+1)+20*y^2*tan(x/2)^4*(1/(tan(x/2)^2+1))^6-32*y^2*tan(x/2)^6*(1/(tan(x/2)^2+1))^6/(tan(x/2)^2+1)-16*tan(x/2)^4*(1/(tan(x/2)^2+1))^6-8*y^2*tan(x/2)^3*(1/(tan(x/2)^2+1))^5+16*tan(x/2)^6*(1/(tan(x/2)^2+1))^6/(tan(x/2)^2+1)-8*y^2*tan(x/2)^5*(1/(tan(x/2)^2+1))^5/(tan(x/2)^2+1)-16*tan(x/2)^3*(1/(tan(x/2)^2+1))^5-4*y^2*tan(x/2)^2*(1/(tan(x/2)^2+1))^4+16*tan(x/2)^5*(1/(tan(x/2)^2+1))^5/(tan(x/2)^2+1)-4*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)+(1/(tan(x/2)^2+1))^2-tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4*tan(x/2)*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1))/(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1))-8*y*tan(x/2)/((tan(x/2)^4+2*tan(x/2)^2+1)*(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1)))-8*y*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/((tan(x/2)^2+1)*(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1)))+16*y*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/((tan(x/2)^2+1)*(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1)))+8*y*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1))+16*y*tan(x/2)^4*(1/(tan(x/2)^2+1))^4/(8*tan(x/2)/(tan(x/2)^4+2*tan(x/2)^2+1)-32*tan(x/2)^3*(1/(tan(x/2)^2+1))^3/(tan(x/2)^2+1)-16*tan(x/2)^2*(1/(tan(x/2)^2+1))^2/(tan(x/2)^2+1)+4/(tan(x/2)^2+1))+y;solve([diff(f(x,y),x)=0,diff(f(x,y),y)=0],[x,y]).

Wolfram Mathematica 11.3 is also not ideal.But in fact this equation system does have an exact solution. Wolfram Mathematica's code

Code:

Solve[D[((

       Sec[x/2]^8 Sqrt[

        Cos[x/2]^12 (2 - y^2 + y^2 Cos[2 x] + 2 Sin[x]) (-1 + 

           Tan[x/2])^2 (1 + Tan[x/2])^4])/Sqrt[2] + 

       y (1 + Tan[x/2])^2 (1 - 2 Tan[x/2] + Tan[x/2]^4))/((-1 + 

         Tan[x/2])^2 (1 + Tan[x/2])^4), x] == 0 && 

  D[((Sec[x/2]^8 Sqrt[

        Cos[x/2]^12 (2 - y^2 + y^2 Cos[2 x] + 2 Sin[x]) (-1 + 

           Tan[x/2])^2 (1 + Tan[x/2])^4])/Sqrt[2] + 

       y (1 + Tan[x/2])^2 (1 - 2 Tan[x/2] + Tan[x/2]^4))/((-1 + 

         Tan[x/2])^2 (1 + Tan[x/2])^4), y] == 0, {x, y}, Reals]