Bug Description

math.setUnion() produces incorrect results whenever the two input sets share any elements. The intersection is placed at the end of the output rather than in its correct sorted position.

Reproduction

math.setUnion([1, 2, 3, 4], [3, 4, 5, 6])
// actual:   [1, 2, 5, 6, 3, 4]
// expected: [1, 2, 3, 4, 5, 6]  ← as documented in the JSDoc

math.setUnion([1, 2], [2, 3])
// actual:   [1, 3, 2]
// expected: [1, 2, 3]

The JSDoc example explicitly documents setUnion([1, 2, 3, 4], [3, 4, 5, 6]) → [1, 2, 3, 4, 5, 6].

Root Cause

The implementation uses:

return concat(setSymDifference(b1, b2), setIntersect(b1, b2))

setSymDifference returns [a-only elements, b-only elements], so the result is
[a-only, b-only, a∩b] — the intersection lands at the end instead of between the two halves.

Expected Behaviour

Union should return the sorted result: a-only elements ++ common elements ++ b-only elements, i.e.:

return concat(concat(setDifference(b1, b2), setIntersect(b1, b2)), setDifference(b2, b1))

Environment

• mathjs version: develop branch