Re: [Maxima-discuss] Feature request: step by step, an error & not simplified...

[Jochen Z. <zie...@gm...>]

Some extremely un-complicated software which quite often helped me a lot
was just printing intermediate results.
Jochen


On Friday, January 16, 2026, Jérémy Bézairie <jer...@gm...>
wrote:

> Stavros Macrakis do you mean https://maxima.sourceforge.io/
> docs/manual/maxima_singlepage.html#Item_003a-Debugging_002fdeffn_002ftrace
> ?
>
> E.g. inverse a function:
>
> (%i1) assume(0<a,0<b,b<1,0<c,0<=x,x<=%pi);
> y(x):=(a^(1/2)*(1-b*cos(x))^2)/c^(3/2);
> trace(all);
> solve(y=y(x),x);
> (%o1)           [a > 0, b > 0, b < 1, c > 0, x >= 0, x <= %pi]
> (%i2)
>                                   1/2               2
>                                  a    (1 - b cos(x))
> (%o2)                    y(x) := ────────────────────
>                                           3/2
>                                          c
> (%i3)
> (%o3)                                 [y]
> (%i4)
> 1 Call   y [x]
>                                  2
>            sqrt(a) (1 - b cos(x))
> 1 Return y ───────────────────────
>                      3/2
>                     c
>
> solve: using arc-trig functions to get a solution.
> Some solutions will be lost.
>                          3/4                         3/4
>                         c    sqrt(y)   1            c    sqrt(y)   1
> (%o4)   [x = %pi - acos(──────────── - ─), x = acos(──────────── + ─)]
>                             1/4        b                1/4        b
>                            a    b                      a    b
>
> The 2nd result x is an error: it assumes the step
> 1-b*cos(x)=±y^(1/2)*c^(3/4)/a^(1/4)
> but it is only
> 1-b*cos(x)=y^(1/2)*c^(3/4)/a^(1/4)
>
> The 1st result x is true but not simplified: 2 instances of b & that
> complication with pi, the simplified result:
> x=acos((1-y^(1/2)*c^(3/4)/a^(1/4))/b)
>
> Although trace(all); no step (with 1 wrong)...
>
> I dream of e.g.:
>
> step(all);
> assume(0<a,0<b,b<1,0<c,0<=x,x<=%pi);
> solve((a^(1/2)*(1-b*cos(x))^2)/c^(3/2)=y,x);
>
> (1-b*cos(x))^2=y*c^(3/2)/a^(1/2)
> 1-b*cos(x)=y^(1/2)*c^(3/4)/a^(1/4)
> b*cos(x=1-y^(1/2)*c^(3/4)/a^(1/4)
> cos(x)=(1-y^(1/2)*c^(3/4)/a^(1/4))/b
> x=acos((1-y^(1/2)*c^(3/4)/a^(1/4))/b)
>
> Dear developers, you have work...
>
> So now, step by step would be nice for verification... & also what I mean
> in my 1st email are:
> I like not to get a result without knowing where it comes from, without
> step by step,
> & you can still learn from a complicated software, e.g. a tip we would not
> have thought of.
>
>

-- 

https://jochen-ziegenbalg.github.io/materialien/