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Why is it impossible to determine the range of a in this universally true problem?

Bill Blair
Posted 1 month ago 
Reduce[ForAll[x, 0 < x < \[Pi]/2, 2 a Sin[a x] >= 0], a, Reals]

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POSTED BY: Bill Blair
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Bill Blair
Bill Blair
Posted 1 month ago

Thanks for your reply.
POSTED BY: Bill Blair
Bill Blair
Bill Blair
Posted 1 month ago

Thanks for your reply. Your answer gets to the heart of the issue, explaining the underlying algorithm and its scope of application in the code implementation. The equivalent expansion of the last line of code gives the final desired result
POSTED BY: Bill Blair

Removing ForAll from inside Reduce leads to a clear answer:

Reduce[0 < x < \[Pi]/2 &&
  2 a Sin[a x] >= 0 &&
  0 <= a x <= 2 Pi,
 {a, x}]

You can reintroduce ForAll to scan the answer:

red = Reduce[0 < x < \[Pi]/2 &&
   2 a Sin[a x] >= 0 &&
   0 <= a x <= 2 Pi,
  {a, x}]
Map[Reduce[ForAll[x, 0 < x < \[Pi]/2, #]] &,
 List @@ red]
POSTED BY: Gianluca Gorni

Probably the Reduce function (and other functions that use quantifiers, like ForAll and Exists) are based on an algorithm called Cylindrical Algebraic Decomposition (CAD). This algorithm is absolutely brilliant, but it has one huge drawback: it only works with polynomials.

POSTED BY: Mariusz Iwaniuk
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