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If the calculator is powerful enough (as in the case of Mathematica) to evaluate Taylor series, for example the Maclaurin series for e^x when x=0, then in doing
by xamuel 13y ago
If the calculator is powerful enough (as in the case of Mathematica) to evaluate Taylor series, for example the Maclaurin series for e^x when x=0, then in doing so it implicitly admits 0^0=1. If it simultaneously says 0^0 is not 1, then the calculator is inconsistent. (Mathematica IS inconsistent in this example)