AngouriMath
AscendingEndpoints(AngouriMath.Entity,AngouriMath.Entity)
Method (no overloads)
Summary
The two roots as (smaller, larger).
Remarks
SortRealsAndNonReals(System.Collections.Generic.IEnumerable{AngouriMath.Entity}) orders the roots it can compare
and leaves the rest in the order the quadratic formula produced them, which is
(-b - sqrt(D))/(2a) before (-b + sqrt(D))/(2a) -- ascending only
whilea is positive. Solving expr <= 0 negates the expression and
so arrives here with a negative leading coefficient, which is how
(x - a)(x + a) <= 0 came to be answered with an interval running from
|a| to -|a| : empty, with the whole solution set lost.
abs rather than as a case split on the sign of the symbol,
because the ordering has a closed form --min(p, q) is
(p + q - |p - q|)/2 and max(p, q) is (p + q + |p - q|)/2 . So
the answer is the single interval[-|a|; |a|] , which is right for either
sign ofa and for a = 0 , rather than three branches each holding on
part of the line. It also keeps the result a Set: a piecewise is an
Entity and not one, so returning it here is not open in any case.
they are handed back untouched so that nothing a concrete coefficient produces
changes shape.
#757
and leaves the rest in the order the quadratic formula produced them, which is
while
so arrives here with a negative leading coefficient, which is how
because the ordering has a closed form --
the answer is the single interval
sign of
part of the line. It also keeps the result a Set: a piecewise is an
Entity and not one, so returning it here is not open in any case.
they are handed back untouched so that nothing a concrete coefficient produces
changes shape.
#757
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