Positively A Function

The function f:R+R+ f: \mathbb{R}^+ \rightarrow \mathbb{R}^+ satisfies the functional equation

f(f(x))=6xf(x). f(f(x) ) = 6x - f(x).

Determine all such functions f(x) f(x) .
It is clear that f(x)=2x f(x) = 2x is a solution. Are there any others? Why, or why not?

Note: You should not assume that the function is a polynomial, or even that it is continuous.


This expands on a problem posed by Sanjeet

Note by Calvin Lin
4 years, 11 months ago

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Fix some a>0a > 0 and let n=f(a)an = \frac{f(a)}{a}, so that f(a)=naf(a) = na. Then by definition of ff, we must have n0n \ge 0. For kNk \in \mathbb{N}, we denote kk applications of ff to aa by writing fk(a)f^k (a) (i.e. f1(a)=f(a)f^1 (a) = f(a), f2(a)=f(f(a))f^2 (a) = f(f(a)), etc.)

It's easy to show, by induction, that for all kNk \in \mathbb{N}, we have: fk(a)=15((3)k(2n)+2k(n+3))af^k (a) = \frac{1}{5} \left((-3)^k(2-n) + 2^k(n+3) \right) a

Suppose that 0n<20 \le n < 2. By the Archimedean Principle, we can find some kNk \in \mathbb{N} such that k>log3/2(n+32n)k > log_{3/2} \left(\frac{n+3}{2-n} \right) and such that kk is odd. For this value of kk, we have (3)k(2n)+2k(n+3)<0(-3)^k(2-n) + 2^k(n+3) < 0, so that fk(a)<0f^k (a) < 0. This is a contradiction since the range of ff is positive numbers.

Now suppose that n>2n > 2. By the Archimedean Principle, we can find some kNk \in \mathbb{N} such that k>log3/2(n+3n2)k > log_{3/2} \left(\frac{n+3}{n-2} \right) and such that kk is even. For this value of kk, we have (3)k(2n)+2k(n+3)<0(-3)^k(2-n) + 2^k(n+3) < 0, so that fk(a)<0f^k (a) < 0. This is a contradiction since the range of ff is positive numbers.

Therefore, we must have n=2n = 2, so f(a)=2af(a) = 2a.

Therefore, if we unfix aa, we get that f(x)=2xf(x) = 2x for all x>0x > 0.

Ariel Gershon - 4 years, 11 months ago

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Hint: Look at the tags.

Calvin Lin Staff - 4 years, 11 months ago

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We let x=a0x=a_0 and fn(x)=anf^n(x)=a_n. Plugging in x=fn(x)x=f^{n}(x) gives an+2=6anan+1a_{n+2}=6a_n-a_{n+1}.

The characteristic polynomial of this recurrence relation is c2+c6=0c^2+c-6=0 which has roots c=2,3c=2,-3. Thus we can represent an=λ1(2)n+λ2(3)na_n=\lambda_1(2)^n+\lambda_2(-3)^n

all we need to do is find a function that satisfies f(λ1(2)n1+λ2(3)n1)=λ1(2)n+λ2(3)nf(\lambda_1(2)^{n-1}+\lambda_2(-3)^{n-1})=\lambda_1(2)^n+\lambda_2(-3)^n.

Now note that (3)n(-3)^n grows faster than (2)n(2)^n, no matter what constants λ1,λ2\lambda_1,\lambda_2 we pick. However, since (3)n(-3)^n oscillates between positive and negative, that would imply that when the function is applied to some positive xx, then f(x)f(x) is negative! Thus, we must have λ2=0\lambda_2=0. Therefore, we want f(λ1(2)n1)=λ1(2)nf(\lambda_1(2)^{n-1})=\lambda_1(2)^{n} Subbing x=λ1(2n1)x=\lambda_1(2^{n-1}), we see that f(x)=2x\boxed{f(x)=2x} which is our only solution.

Does this work?

Daniel Liu - 4 years, 11 months ago

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Nice. A wording correction though. Note that if λ1\lambda_1 is large enough and 0<λ2<10<\lambda_2<1 is very small then the function will give a lot of positive values at first. But as nn gets arbitrarily large it will eventually start to give negative values. So you might want to clarify that. :)

Jubayer Nirjhor - 4 years, 11 months ago

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That's where I'm stuck too. :/

Jubayer Nirjhor - 4 years, 11 months ago

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I think I got it. Edited the post.

Daniel Liu - 4 years, 11 months ago

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Looking at the functional equation, it is apparent that f(x) must be linear and that it cannot have a constant term. Thus f(x)=k*x. To find the values that k can assume, substitute in the functional equation: k^2=k-6; solutions are k=-3 or 2. Since the function is defined in Positive Real space, it has to be f(x)=2x.

Amruta Vasudevan - 4 years, 11 months ago

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Why must it be linear, or even continuous?

Note that if the condition that the range is positive, then infinitely many solutions exist, because we can define f(1)f(1) arbitrarily.

Calvin Lin Staff - 4 years, 11 months ago

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I realize that now. Although, if the function is continuous, am I wrong in presuming that it must also be linear? Given the above equation, I don't see any other scenario.

Amruta Vasudevan - 4 years, 11 months ago

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F(x) equal to -3x

Vaibhav Raghavan - 4 years, 11 months ago

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f(x)=-3x is a solution. I found it by solving the quadratic equation f^2+f-6=0. Do not ask me why :-)

Hatim Zaghloul - 4 years, 11 months ago

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The function's range and domain are both positive numbers. However -3x has a range of negative numbers.

Ariel Gershon - 4 years, 11 months ago

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Thanks. I vaguely recall that solving the quadratic equation gives the solutions. However I am not sure.

Hatim Zaghloul - 4 years, 11 months ago

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I can say that there are no other solutions but I may assume that the function is a polynomial since from the given (somehow confusing logic) that f is a polynomial function. But the assumption...

John Ashley Capellan - 4 years, 11 months ago

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That's my error. I've updated the question. Sorry for the confusion.

Calvin Lin Staff - 4 years, 11 months ago

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I found 2 solutions: f(x) = 2x, and f(x) = -3x. :)

Carlos Bravo - 4 years, 11 months ago

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