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Problems like these (typically restricted to integers/natural numbers) come up in competitions from time to time. Sometimes there are some tricks required that
by dan-robertson 7y ago
Problems like these (typically restricted to integers/natural numbers) come up in competitions from time to time. Sometimes there are some tricks required that are very hard to find on the spot. Here is a related problem you may enjoy:
Find all increasing functions f : N -> N such that f(f(x)) = 3x
- fyp 7y agoIs this problem correctly stated? Substituting in x=f(y) you get f(f(f(y))) = 3f(y) so f(3y) = 3f(y) This suggests a linear solution but then you get f(x)=sqrt(3)x which doesn't work since it's N -> N.
- Chinjut 7y agoSuggests is not demands. If we discard the requirement to be an increasing function, one recursively defined solution takes f(3x + 1) = 3x + 2, f(3x + 2) = 3 * (3x + 1), and f(3x) = 3f(x), with f(0) = 0. Or analogously for any way of splitting the positive integers not divisible by 3 into ordered pairs (a, b). Then define f(3^n a) = 3^n b and f(3^n b) = 3^(n + 1) a. This is the general form of the solutions.
- a1369209993 7y ago∅ assume f exists increasing requires that x < f(x) < f(f(x)) f(f(0)) = 3*0 = 0 ≮ 0 therefore f does not exist, qed
- AnimalMuppet 7y agof(x) = sqrt(3) * x. Then f(f(x)) = sqrt(3) * sqrt(3) * x = 3x. So there is at least one solution.
- a1369209993 7y agoThat's not increasing (√3 * 0 = 0 ≮ 0) and it's not on natural numbers (√3 * 1 = √3 (≈ 1.732) ∉ ℕ).
- JoshuaDavid 7y agoI 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".
- a1369209993 7y agoIgnoring zero, I get the summation, starting at [1]=2, of {1,3,1,1,1,3,3,3,1,1,1,1,1,1,1,1,1,3,...} = {..{1}^3^i,..{3}^3^i for i in ℕ} Edit: derp, it's ℕ, not ℤ⁺.
- dan-robertson 7y agoThe definition of an increasing function is a function f such that for all x and y, x < y implies f(x) <= f(y). For a strictly increasing function change <= to <. Your argument fails because you drop the antecedent in: x < f(x) => f(x) < f(f(x)). If f(0) = 0, you don’t have 0 < f(0) so can’t require that f(f(0))<f(0). Note however that it is easy to show that f must be strictly increasing if it is increasing for if x is not y and f(x) = f(y) then 3x = f(f(x)) = f(f(y)) = 3y which cannot be the case.
- sgdpk 7y agoInteresting! The form of the solution can be written conveniently in base 3.