4 ms·
I think in this context "increasing" means f(x+ε)>f(x) for all x, not f(x) > x for all x.
by JoshuaDavid 7y ago
I think in this context "increasing" means f(x+ε)>f(x) for all x, not f(x) > x for all x.
- a1369209993 7y agoWhat is ε? The original problem is about natural numbers (ie nonnegative/unsigned integers).
- AnimalMuppet 7y agoWell then, f(x + 1) > f(x). I don't know where you're getting this x < f(x) as "increasing". The point about sqrt(3) not working for the naturals is a valid point, however.
- a1369209993 7y ago> Well then, f(x + 1) > f(x). I don't know where you're getting this x < f(x) as "increasing". Headdesk Yes, that's correct; I misparsed "increasing functions f" as "functions f that increase [the value passed through them]" and then got distracted by the discrete vs continuous confusion.
- AnimalMuppet 7y agoWell, I missed that the domain was N, so... let's both just blame the tryptophan and call it good.
- JoshuaDavid 7y agoε is epsilon, conventionally "an arbitrarily small positive number".