Comments in line.
On Sat, Mar 30, 2013 at 10:51 PM, Thomas D. Dean <tomdean at speakeasy.org>wrote:
> eq:2*x^3-x^2*y+y^3-1=0; p:[2,-3];
>
> eq1:''rhs(first(solve(diff(eq,**x),diff(y,x))));
>
^^^ Presumably there is a depends(y,x) before this. Or use 'diff(y,x).
^^^ Why are you using [''] (two single quotes) here? In this case, it has
no effect at all.
^^^ Why are you looking at the first solution only? The order of the
solutions is arbitrary.
> df(x,y):=''eq1;
>
^^^ This is an appropriate use of [''], though I'd recommend define
instead, since it works equally well in programmatic cases.
> pointslope(p,m):=y-p[2]=m*(x-**p[1]);
> eq2:pointslope(p,df(p[1],p[2])**);
> soln:rhs(first(solve(eq2,y)));
>
> eq3:last(factor(solve(eq,y))); /* choose the real solution */
> plot2d([rhs(eq3),soln],[x,-4,**4],[y,-10,10]);
>
> Here is my problem.
>
> eq3:factor(solve(eq,y));
> eq has three solutions for y. How do I choose the right one? I used the
> if statement for simpler results. But, I cannot reduce this to equality.
>
> I have been searching for a way to tell if any parts of eq3 are real.
>
Better to do the solve after substituting specific values, rather that
substituting values into the result. The result is more likely to be
tractable that way. Then you can check whether imagpart(root)=0. In this
case, the roots are messy; you'll want imagpart(radcan(root)).
float(subst([x=p[1],y=p[2]],**eq3[1]));
^^^ apparently this refers to your second eq3. It would have been
clearer if you'd used different variables
float(subst([x=p[1],y=p[2]],**eq3[2]));
> float(subst([x=p[1],y=p[2]],**eq3[3]));
>
> if subst([x=p[1],y=p[2]],eq3[1]) then eq4:eq3[1]
>
>
elseif subst([x=p[1],y=p[2]],eq3[2]) then eq4:eq3[2]
> elseif subst([x=p[1],y=p[2]],eq3[3]) then eq4:eq3[3];
>
> Is there a reasonable way to do this?
>
> Tom Dean
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