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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 $x > 0$.

Log in to reply

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?

Log in to reply

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

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

Log in to reply

I think I got it. Edited the post.

Log 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.

Log in to reply

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.

Log in to reply

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) equal to -3x

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 :-)

Log in to reply

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

Log in to reply

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

Log in to reply

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...

Log in to reply

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

Log in to reply

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

Log in to reply