#hey guys I just proved x^infinity =0
6 messages · Page 1 of 1 (latest)
6 messages · Page 1 of 1 (latest)
in the proof you assume such an alpha exists
what you are doing is basically assuming |x|<1
in which case it is what we expect, as multiplying by x gives something smaller
But yea other values of alpha exist
For x = 1 alpha can be any number