I am trying to find a function $f(x,y,z)$,polynomial with respect to x , y and z such that if I knew $f(x,y,z)$ I could find $x,y,z$
Asked
Active
Viewed 37 times
-1
-
1Do you need $f$ to be continuous? Do you have any limitations for $x,y$ and $z$ ? Do you need $f\in \mathbb{R}^1$, as opposed to ${}^2$ or ${}^3$ ... – Matti P. Sep 22 '23 at 12:14
-
1How about $f(x,y,z) = x$ – messenger Sep 22 '23 at 12:17
-
@messenger no all x,y,z must be included. – Cerise Sep 22 '23 at 13:40
-
@MattiP. $x,y,z>0$ and $f$ doesnt need to be continuous.Also $f(x,y,z)$ must give at the output a integer only. – Cerise Sep 22 '23 at 13:41
1 Answers
1
You are looking for an injective polynomial from $\mathbb{R}^3\to\mathbb{R}$. Polynomials are (among other things) continous and unfortunately, there exists not even any continous injective function from $\mathbb{R}^3\to\mathbb{R}$. Look here for a reference: Existence of an injective continuous function $\Bbb R^2\to\Bbb R$?
If you drop the continuity (and therefore also the polynomial nature) you can achieve such things, are detailed post can be found here: Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$
Another option as suggested by Matti P. would be to restrict your domain to a subset $\Omega\subset\mathbb{R^3}$ i.e. by a curve $\Omega=\{(\gamma_x(t),\gamma_y(t),\gamma_z(t))\in\mathbb{R^3}|t\in\mathbb{R}\}$. This could allow you to find such a polynomial.
Finally, if you let you "polynomial" have values in $\mathbb{R^3}$ there exists many solutions the easiest of course being $f(x,y,z)=(x,y,z)$ (but I dont think that is what you had in mind when you posted your question).
maxmilgram
- 3,735