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For u = -1 in the polynomial shape we end up with a 0/0 expression which evaluates to NaN. However, we know from l'Hopital's rule that the result should beo 0. This PR intervenes for -1 to yield 0. WHile at it, since this function and its gradietn are only defined over ±1 I check r and set results to 0 if outside.
maxrjones
approved these changes
Mar 23, 2022
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For u = -1 in the polynomial shape we end up with a 0/0 expression which evaluates to NaN. However, we know from l'Hopital's rule that the result should be 0. This PR intervenes for -1 to yield 0. While at it, since this function and its gradient are only defined over ±1 I check r and set results to 0 if outside.