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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## Comments

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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 \(x > 0\).

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Hint:Look at the tags.Log in to reply

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?

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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. :)

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

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I think I got it. Edited the post.

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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.

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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 \(f(1)\) arbitrarily.

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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.

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F(x) equal to -3x

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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 :-)

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

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

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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...

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

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

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