We CANNOT use FPTaylor to get the Symbolic Error Form

[rospoly]

The reason is FPTaylor handles any non-linearity using a worst-case approach. Both first-order Taylor expansion and symbolic affine arithmetic substitute any non-linearity with a numeric interval. In FPTaylor, since the worst-case approach, this interval is the result of a maximization problem with NO constraints. On the other hand, in PAF, we handle any non-linearity probabilistically , thus the numeric interval is the result of a maximization problem with some constraints given by the confidence interval. This has an impressive impact when the support of the input variables is huge (e.g. Exponential(0, MAX_DOUBLE)). The immediate consequence is PAF ends up with a symbolic error form that is different from the one in FPTaylor.