Instead of $x^{4/3}$ do $(x^4)^{1/3}$

Question

I'm doing a little filter processing research and at one point I expand a signal by applying: $x^{4/3}$ to the signal which will only yield real outputs for real, positive inputs. Since $(x^a)^b = (x^b)^a $ and $(x^a)^b = x^{a \cdot b}$ I asked myself why can't I calculate $x^{4/3}$ by calculating $(x^4)^{1/3}$ of which $x^4$ will only yield positive values ? This way one could plug in negative values into the function? Would this function still correctly reverse $x = z^{3/4}$ ?

EDIT: I want to achieve only real number outputs, that's the reason for this whole thing

Answer 1

When $x>0,$ there should be no issue. On some calculators you will have an issue doing $(x^4)^{1/3}$ versus $(x^{1/3})^4,$ because some calculators use logarithms to compute complex numbers.

In other words, your procedure is a good one to avoid running into complex numbers. I will say also that some calculators, particularly newer ones, don't have this problem to begin with and will calculate the answer correctly either way.

For $x=z^{3/4},$ if you want real outputs, then $z\ge 0,$ period. This procedure will not work. The difference is that fundamentally, with the first case, you're taking an odd root, which is defined for all real numbers. In the second case, you're taking an even root, which is only real-valued for non-negative numbers.

Answer 2

To work around the sign issue, use odd functions:

$$f(x)=\begin{cases}x\ge0\to x^{4/3},\\x\le0\to -(-x)^{4/3},\end{cases}$$ and

$$f^{-1}(x)=\begin{cases}x\ge0\to x^{3/4},\\x\le0\to -(-x)^{3/4}.\end{cases}$$

But from your description, the fact that the filter returns negative values meant to be powers seems to reveal a methological problem (this compression/filtering/decompression procedure might be inappropriate).