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V. Timofte-;-?-{0, .. .. 1}-;, .. .. , ,. .. {0, and .. , z ? k+1 } and so is ?-invariant. Using our earlier argument we see that for i in {? k + 1, . . . , ? k + s} the divided difference q {i,? k +s+1,...,?r} is ?-invariant. As a result, p ? k +j itself is ?-invariant. It remains to prove that p ? k +j is ?-invariant for a permutation ? of {? k + 1, . . . , ? k+1 }. We do this first for ? = (? k + 1, ? k + 2), by proving that all summands in the definition of p ? k +j are ?-invariant. For any s in {2, . . . , j}, ? j?s (z ? k +s+2 , . . . , z ? k +1 ) does not depend on (z ? k +1 , z ? k +2 ), so it is ?-invariant, The fact that all entries of p are polynomials follows from our first assumption. Proving that they are S ? -invariant requires more work, as we have to deal with numerous cases. While most are straightforward, the last case does involve nontrivial calculations. Fix k, vol.284, pp.174-190, 2003.