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Positively A Function

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

Note by Calvin Lin
3 years, 1 month ago

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Fix 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 · 3 years, 1 month 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 · 3 years, 1 month ago

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@Daniel Liu 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. :) Jubayer Nirjhor · 3 years, 1 month ago

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@Daniel Liu That's where I'm stuck too. :/ Jubayer Nirjhor · 3 years, 1 month ago

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@Jubayer Nirjhor I think I got it. Edited the post. Daniel Liu · 3 years, 1 month ago

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Hint: Look at the tags. Calvin Lin Staff · 3 years, 1 month 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 · 3 years ago

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@Amruta Vasudevan 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. Calvin Lin Staff · 3 years ago

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@Calvin Lin 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 · 3 years 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 · 3 years, 1 month ago

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@Hatim Zaghloul The function's range and domain are both positive numbers. However -3x has a range of negative numbers. Ariel Gershon · 3 years, 1 month ago

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@Ariel Gershon Thanks. I vaguely recall that solving the quadratic equation gives the solutions. However I am not sure. Hatim Zaghloul · 3 years, 1 month ago

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F(x) equal to -3x Vaibhav Raghavan · 3 years, 1 month ago

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Comment deleted Jun 23, 2014

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@Jubayer Nirjhor Calvin said you can't assume the function is a polynomial :/ Bogdan Simeonov · 3 years, 1 month ago

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Comment deleted Jun 23, 2014

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@Jubayer Nirjhor "You may not" is a nice phrase for "whatever you do, you absolutely should never ever ever ..." Calvin Lin Staff · 3 years, 1 month ago

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@Calvin Lin Sorry, I wasn't aware. Jubayer Nirjhor · 3 years, 1 month ago

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@Jubayer Nirjhor 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. Calvin Lin Staff · 3 years, 1 month ago

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@Calvin Lin Actually I considered finding the non-polynomial solution as an extra job. Also you wrote

The polynomial function \(f:\mathbb{R}^+\to\mathbb{R}^+\) satisfies ........ ........

so it's easy to be fooled. :) Jubayer Nirjhor · 3 years, 1 month ago

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@Jubayer Nirjhor Ah yes, that's my error. Updated. Sorry for the confusion. Calvin Lin Staff · 3 years, 1 month ago

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@Calvin Lin 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. Daniel Liu · 3 years, 1 month ago

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@Daniel Liu Exactly. Jubayer Nirjhor · 3 years, 1 month ago

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I found 2 solutions: f(x) = 2x, and f(x) = -3x. :) Carlos Bravo · 3 years 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 · 3 years, 1 month ago

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@John Ashley Capellan That's my error. I've updated the question. Sorry for the confusion. Calvin Lin Staff · 3 years, 1 month ago

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