adding   .sage() does not help as in my example?
Please, post a minimal example, if possible.

R.

On 8 říj, 18:31, Mikie <[email protected]> wrote:
> Here is the example that is the problem.  Your last suggestion did
> solve the problem with maxima.solve.  But now how do I stop the other
> \over.
>
> -----------------------------------------------
> def CSquare(r1):
>         eq11=r1;y=b*1
>         w1=maxima.subst(eq11,b,y)
>         w2 = latex(eq11); w5 = "$"+w2+"=0"+"$"
>         V1=p/2;RHalf=maxima.coeff(V1,p,1);#this is 1/2
>         LSide1a=maxima.lhs(w1)
>         L1= maxima.coeff(LSide1a,x,1)#coef of linear term
>         L2= maxima.coeff(LSide1a,x,0)#constant
>         L3= maxima.coeff(LSide1a,x**2,1)#constant
>         eq2=factor(w1)
>         eq4=(1/L3)*LSide1a#mult by coef of quad
>         eq10=expand(eq4)#div out the constant
>         #print eq10
>         R2 = latex(eq10);R3="$"+R2+"$"
>         M1=maxima.args(eq10);L10=len(M1)
>         Cof1=maxima.args(eq10);Cof1a=Cof1[1];Cof1b=Cof1[2]
>         Const = Cof1[2] #constant
>         Half2=maxima.coeff(Cof1a,x,1);
>         val2=(RHalf*(Half2))**2 #Addon
>         Addon=maxima.ev(val2*val2)
>         eq5=M1[0]+M1[1]+val2
>         Rside = -Cof1b+val2
>         eq22=eq5;z=b*1==Cof1b
>         w3=maxima.subst(eq5,b,z)
>         eq6=factor(eq5)
>         w4=maxima.subst(eq6,b,z)
>         LeftS=maxima.lhs(eq6+Rside)
>         eq7=Cof1a
>         eq8=Rside
>         Soln=solve(w1)
>         return w5,R3,eq5,eq22,eq6,Rside,Soln
> -------------------------------------------------------------
> Here is the output
> '${x}^{2}  - x - \\frac{3}{2}=0$'
> '$x^2-x-{{3}\\over{2}}$'<<<<<<<<< the problem
> x^2-x+1/4
> x^2-x+1/4
> (2*x-1)^2/4
> 7/4
> [x=-(sqrt(7)-1)/2,x=(sqrt(7)+1)/2]
> --------------------------------------------------
> Input --val1=CSquare(x^2-x-3/2);val1[0];val1[1];val1[2];val1[3];val1
> [4];val1[5];val1[6]
>
> On Oct 6, 3:51 pm, "[email protected]" <[email protected]> wrote:
>
> > This \over is from Maxima.
> > try
>
> > a1=maxima.solve(x^2-x-3,x)
> > R1=a1[0]
> > R3=maxima.rhs(R1).sage()
> > latex(R3)
>
> > R.
>
> > On 6 říj, 22:25, Mikie <[email protected]> wrote:
>
> > > Yes, your right.  I am latexing a value from maxima.solve(x^2-x-3,x).
> > > Then maxima.rhs().  Then latexing the value.  It still gives me the
> > > \over.  I am using 3.4.
> > > When I assign it to a variable it works as below.  If you would try
> > > a1=maxima.solve(x^2-x-3,x)
> > > R1=a1[0]
> > > R3=maxima.rhs[R1]
> > > latex(R3)
> > > produces -- {{1-\sqrt{13}}\over{2}}
> > > not good
>
> > > On Oct 6, 1:46 pm, "[email protected]" <[email protected]> wrote:
>
> > > > Hm, this is my Sage 4.1.1
>
> > > > a1=-(sqrt(13)-1)/2
> > > > latex(a1)
>
> > > > output is -\frac{1}{2} \, \sqrt{13} + \frac{1}{2}
>
> > > > You may have old version of Sage
>
> > > > latex(-{{\sqrt[13]-1\over[2}}) produces error
>
> > > > Robert Marik
>
> > > > On 6 říj, 21:10, Mikie <[email protected]> wrote:
>
> > > > > If I have this a1=-(sqrt(13)-1)/2 in a variable, then latex(a1) it
> > > > > produces -{{\sqrt[13]-1\over[2}}.
> > > > > If I do  latex(-{{\sqrt[13]-1\over[2}}) get \frac{1-\sqrt{13}}{2},
> > > > > which is what I want.
>
> > > > > This is in a function.  I need the latter.  The \over  does not do the
> > > > > pretty print.
> > > > > Is there a work around?
> > > > > Thanx
>
> > > > > ------------------------------------
> > > > > def Solver2a(exp1,exp2):
> > > > >         eq11=exp1;w=b*1==0
> > > > >         w1=maxima.subst(eq11,b,w)
> > > > >         R1 = maxima.solve(exp1,exp2);
> > > > >         w2 = latex(eq11); w3 = "$"+w2+"$"
> > > > >         R2 = latex(R1);R3="$"+R2+"$"
> > > > >         R4 = R1[0];R5=maxima.rhs(R4);R6=latex(R5);
> > > > >         str1="Calculate the symbolic solutions for the following
> > > > > equation."
> > > > >         str2="And here is the solution :"
> > > > >         return str1,w3,str2,R1,R6,R5
> > > > > -------------------------------------- Hide quoted text -
>
> > > > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -
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