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Mathematics
Topology and Analysis
Higher Order Derivative Test and Germs
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[QUOTE="andrewkirk, post: 6044833, member: 265790"] Yes, that follows from the assumption '##f^{(0)}=0## for all [natural] ##n##' if we take the usual meaning of natural numbers to be all non-negative integers, since ##f^{(0)}=f##. If we take the alternative interpretation that natural numbers exclude 0, the reasoning doesn't change. We just need to replace range values of 0 by ##f(0)##. I doubt whether such a function is possible, because I think the domain of the function may have a higher cardinality than the range, in which case an injective function is impossible. [/QUOTE]
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Higher Order Derivative Test and Germs
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