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

\[ f(f(x) ) = 6x - f(x). \]

Determine all such functions \( f(x) \).

It is clear that \( f(x) = 2x \) 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

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TopNewestFix some \(a > 0\) and let \(n = \frac{f(a)}{a}\), so that \(f(a) = na\). Then by definition of \(f\), we must have \(n \ge 0\). For \(k \in \mathbb{N}\), we denote \(k\) applications of \(f\) to \(a\) by writing \(f^k (a)\) (i.e. \(f^1 (a) = f(a)\), \(f^2 (a) = f(f(a))\), etc.)

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

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

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

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

Therefore, if we unfix \(a\), we get that \(f(x) = 2x\) for all \(x > 0\). – Ariel Gershon · 2 years, 11 months ago

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

The characteristic polynomial of this recurrence relation is \(c^2+c-6=0\) which has roots \(c=2,-3\). Thus we can represent \[a_n=\lambda_1(2)^n+\lambda_2(-3)^n\]

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

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

Does this work? – Daniel Liu · 2 years, 11 months ago

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– Jubayer Nirjhor · 2 years, 11 months ago

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

– Jubayer Nirjhor · 2 years, 11 months ago

That's where I'm stuck too. :/Log in to reply

– Daniel Liu · 2 years, 11 months ago

I think I got it. Edited the post.Log in to reply

Hint:Look at the tags. – Calvin Lin Staff · 2 years, 11 months agoLog in to reply

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 · 2 years, 11 months ago

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Note that if the condition that the range is positive, then infinitely many solutions exist, because we can define \(f(1)\) arbitrarily. – Calvin Lin Staff · 2 years, 11 months ago

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– Amruta Vasudevan · 2 years, 11 months ago

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.Log in to reply

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 · 2 years, 11 months ago

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– Ariel Gershon · 2 years, 11 months ago

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

– Hatim Zaghloul · 2 years, 11 months ago

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

F(x) equal to -3x – Vaibhav Raghavan · 2 years, 11 months ago

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– Bogdan Simeonov · 2 years, 11 months ago

Calvin said you can't assume the function is a polynomial :/Log in to reply

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– Calvin Lin Staff · 2 years, 11 months ago

"You may not" is a nice phrase for "whatever you do, you absolutely should never ever ever ..."Log in to reply

– Jubayer Nirjhor · 2 years, 11 months ago

Sorry, I wasn't aware.Log in to reply

– Calvin Lin Staff · 2 years, 11 months ago

No worries. That's the reason why I set this as Level 4, because it would be a common misconception to merely solve this question over the polynomials, and not consider any other types of functions.Log in to reply

so it's easy to be fooled. :) – Jubayer Nirjhor · 2 years, 11 months ago

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– Calvin Lin Staff · 2 years, 11 months ago

Ah yes, that's my error. Updated. Sorry for the confusion.Log in to reply

– Daniel Liu · 2 years, 11 months ago

perhaps he was showing that \(f(x)=2x\) was the only polynomial that satisfies it, and he was letting the general case be solved by other people.Log in to reply

– Jubayer Nirjhor · 2 years, 11 months ago

Exactly.Log in to reply

I found 2 solutions: f(x) = 2x, and f(x) = -3x. :) – Carlos Bravo · 2 years, 10 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 · 2 years, 11 months ago

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– Calvin Lin Staff · 2 years, 11 months ago

That's my error. I've updated the question. Sorry for the confusion.Log in to reply