Re: turning differences into sums

Try

factor(1/ratsimp(1/ex)),algebraic;

ratsimp/algebraic generally eliminates roots in the denominator

or more "manually"

apply("/",factor(expand(args(ex)*(sqrt(a)+sqrt(b)))))


On Sun, Jun 21, 2026, 04:46 Claude Heiland-Allen <[email protected]> wrote:

> Hi all,
>
> I have something like
>
> (sqrt(a) - sqrt(b)) / c
>
> and I would like to multiply to and bottom by
>
> (sqrt(a) + sqrt(b))
>
> and simplify each to get
>
> (a - b) / (c * (sqrt(a) + sqrt(b)))
>
> of course my actual expressions are much more complicated (and I hope my
> (a-b) will simplify further) and I cannot see how to do it programmatically
> without lots of manual copy and paste with potential for errors.
>
> actual code:
>
> s(x,y,z,w) := (x^2 + y^2 - z^2 - w^2) * (x^2 - y^2) / (x^2 + y^2);
> t(x,y,z,w) := (x^2 + y^2 - z^2 - w^2) * (2 * x * y) / (x^2 + y^2);
> u(x,y,z,w) := 2 * sqrt((x^2 + y^2) * (z^2 + w^2)) * (z^2 - w^2) / (z^2 +
> w^2);
> v(x,y,z,w) := 2 * sqrt((x^2 + y^2) * (z^2 + w^2)) * (2 * z * w) / (z^2 +
> w^2);
> p(f) := factor(expand(f(X+x,Y+y,Z+z,W+w) - f(X,Y,Z,W)));
> p(s);
> p(t);
> p(u);
> p(v);
>
> the last two expressions are the ones I want to manipulate as described,
> here is the last one:
>
> -((4*(W*sqrt(Y^2+X^2)*Z*sqrt(z^2+2*Z*z+w^2+2*W*w+Z^2+W^2)
>             -sqrt(Z^2+W^2)*w*sqrt(y^2+2*Y*y+x^2+2*X*x+Y^2+X^2)*z
>             -W*sqrt(Z^2+W^2)*sqrt(y^2+2*Y*y+x^2+2*X*x+Y^2+X^2)*z
>             -Z*sqrt(Z^2+W^2)*w*sqrt(y^2+2*Y*y+x^2+2*X*x+Y^2+X^2)
>             -W*Z*sqrt(Z^2+W^2)*sqrt(y^2+2*Y*y+x^2+2*X*x+Y^2+X^2)))
>  /(sqrt(Z^2+W^2)*sqrt(z^2+2*Z*z+w^2+2*W*w+Z^2+W^2)))
>
>
> context: I'm trying to apply perturbation techniques to iterations of the
> 4D Hopfbrot fractal, a cousin of the famous 3D Mandelbulb.  Perturbation
> techniques popularized in the last years allow computationally efficient
> deep zooming of 2D fractals like the Mandelbrot set, and recently have been
> applied to the Mandelbulb.  The main idea is to use "small" differences
> from one "large" reference orbit, and the key step is symbolically
> simplifying so there is no catastrophic cancellation when evaluating
> numerically (i.e. naively (X+x)-X will give 0 instead of x when x << X).
>
> Thanks,
>
>
> Claude
>
> https://mathr.co.uk
>
>
> _______________________________________________
> Maxima-discuss mailing list
> [email protected]
> https://lists.sourceforge.net/lists/listinfo/maxima-discuss
>

_______________________________________________
Maxima-discuss mailing list
[email protected]
https://lists.sourceforge.net/lists/listinfo/maxima-discuss