# 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
4 years 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 $$x > 0$$.

- 4 years ago

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?

- 4 years 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. :)

- 4 years ago

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

- 4 years ago

I think I got it. Edited the post.

- 4 years ago

Hint: Look at the tags.

Staff - 4 years ago

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.

- 4 years ago

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.

Staff - 4 years 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.

- 4 years ago

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

- 4 years ago

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

- 4 years ago

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

- 4 years ago

F(x) equal to -3x

- 4 years ago

Comment deleted Jun 23, 2014

Calvin said you can't assume the function is a polynomial :/

- 4 years ago

Comment deleted Jun 23, 2014

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

Staff - 4 years ago

Sorry, I wasn't aware.

- 4 years 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.

Staff - 4 years ago

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. :)

- 4 years ago

Ah yes, that's my error. Updated. Sorry for the confusion.

Staff - 4 years 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.

- 4 years ago

Exactly.

- 4 years ago

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

- 4 years ago

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...

- 4 years ago

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

Staff - 4 years ago