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From: Jochen Z. <zie...@gm...> - 2026-01-16 17:39:48
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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/ |