2 ms·
But they don't retain their status as hypernaturals! dx does not need to evenly divide the interval over which the integral is taken. Whenever it doesn't, the n
by thaumasiotes 2y ago
But they don't retain their status as hypernaturals! dx does not need to evenly divide the interval over which the integral is taken. Whenever it doesn't, the number of slices in the integral will fail to be a hypernatural number, because one of the slices will extend beyond the interval boundary.
The theorem tells us that the area of the extended interval that uses a hypernatural number of slices has the same real part as the area of the exact interval. It doesn't tell us that the exact interval contains a hypernatural number of slices.
- yuppiemephisto 2y agoThat's what "up to an infinitesimal error anyway" meant.
- thaumasiotes 2y agoYes, but that's also the entire thing I was questioning. The essay says that an integral necessarily contains a hypernatural number of infinitesimal slices. I don't think that's true.