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

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 8:34
• The domain and range are both the nonnegative reals – Miles Johnson Oct 15 '16 at 22:49

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

• And so is $f(x)=-x^2$. – celtschk Oct 15 '16 at 8:47
• Also $f(x) = -1/x^2$. – Math Student Oct 15 '16 at 8:48
• For those functions we get $f(f(x))=-x^4$ – arberavdullahu Oct 15 '16 at 8:56
• @arberavdullahu you are right. – Jean Marie Oct 15 '16 at 9:02
• $f_1(x)=\begin{cases} -x^2 &\text{for } x \ge 0\\x^2 &\text{for } x<0 \end{cases}, f_2(x)=\begin{cases} -1/x^2 &\text{for } x \ge 0\\1/x^2 &\text{for } x<0 \end{cases}$ are another solutions. – Rafał Oct 15 '16 at 11:23

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.