What solutions are there to $f(f(x))=x^4$?

Question

I know $f(x)=x^2$ is a solution, but I can't seem to find any others and I have no idea how to approach this.

You have do define domain set and range set of f – arberavdullahu Oct 15 '16 at 08:34 — arberavdullahu, Oct 15 '16 at 08:34
The domain and range are both the nonnegative reals – Miles Johnson Oct 15 '16 at 22:49 — Miles Johnson, Oct 15 '16 at 22:49

score 3 · Answer 1 · answered Oct 15 '16 at 08:35

3

$f(x)=1/x^2$ is another one with domain $\mathbb{R^*}$.

answered Oct 15 '16 at 08:35

Jean Marie

81,803

2

And so is $f(x)=-x^2$. – celtschk Oct 15 '16 at 08:47
1

Also $f(x) = -1/x^2$. – keynes Oct 15 '16 at 08:48
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For those functions we get $f(f(x))=-x^4$ – arberavdullahu Oct 15 '16 at 08:56
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@arberavdullahu you are right. – Jean Marie Oct 15 '16 at 09:02
$f_1(x)=\begin{cases} -x^2 &\text{for } x \ge 0\x^2 &\text{for } x<0 \end{cases}, f_2(x)=\begin{cases} -1/x^2 &\text{for } x \ge 0\1/x^2 &\text{for } x<0 \end{cases}$ are another solutions. – Rafał Oct 15 '16 at 11:23

score 2 · Answer 2 · answered Oct 20 '16 at 12:12

This equation leaves us with immense freedom of choice. There is a continuum of continuous solutions, and even more discontinuous ones.

To find a few, draw a freehand increasing graph from $(2,4)$ to $(4,16)$. Then every point in $(4,16)$ is $f(x)$ for some $x\in(2,4)$, so you may define $f(f(x))$ as $x^4$, then continue in a similar manner to define $f$ on $(16,256)$, and so on. Mind you, this does not even cover all continuous solutions, as illustrated by $1\over x^2$.

To find more, split all $\mathbb R_+$ into countable sets $A_x=\{\dots,\sqrt[4]x,x,x^4,x^{16},\dots\}$, then join the sets in pairs $(A_x,A_y)$, select a representative from each set in each pair, and define $f(x)=y$. Then $f(y)=x^4,\;f(x^4)=y^4$, and so on.

What solutions are there to $f(f(x))=x^4$?

2 Answers2