Say we have a nonelementary function $F(x)$ on the real numbers. Let $E_1,E_2,\ldots,E_n$ be a sequence of finite elementary functions on the reals. Is it always true that $$ R(x)=(E_1\circ E_2\circ \cdots\circ E_n\circ F)(x) $$ is also nonelementary? Here we are assuming that $R$ is nontrivial: it cannot be the case that $R$ is any constant, or that any of the $E_i$ are the inverse of $F$ (you get the idea: constant multiples of the inverse, $E_1$ being the inverse of $E_2\circ F$, etc. are also not allowed). I do not have an idea where to start proving or disproving this, but it intuitively seems to be true. Any input is appreciated!
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ What if $E_1$ is the map $x \mapsto x + 1$ and $E_2 \equiv 0$? Then $R(x)$ is not identically $0$, but it also doesn't matter what $F$ is. $\endgroup$– Theo BenditCommented Sep 5, 2018 at 8:24
-
$\begingroup$ Sorry, please see the updated question. $\endgroup$– LucasCommented Sep 5, 2018 at 8:26
-
2$\begingroup$ "You get the idea" The idea I get is that you want all the answers, except the ones you can think of. Unless you properly enumerate these, this isn't a fair question to answer. $\endgroup$– Theo BenditCommented Sep 5, 2018 at 9:05
-
1$\begingroup$ Elementary functions are closed under composition, so you can simplify your question to $R(x) = E(F(x))$. $\endgroup$– PaulCommented Sep 5, 2018 at 11:16
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
No, elementary compositions of nonelementary functions aren't always nonelementary.
My answer is for the elementary functions from Liouville and Ritt. They are defined in differential algebra. That are the functions that are composed of $\exp$, $\ln$ and/or unary or multiary univalued algebraic functions. Take e.g. functions $X\in\mathbb{C}\to Y\in\mathbb{C}$.
see e.g. Wikipedia: Elementary function
$ $
The composition of elementary functions is an elementary function again. Each elementary function can be presented as composition of elementary functions. Therefore you can set
$$E=E_1\circ E_2\circ E_3\circ\ ...\ \circ E_n,$$
with $E$ an elementary function.
$$E(F(x))=R(x)$$
Let $^{-1}$ denote the left or right inverse.
$ $
1.)
Let $G$ be an elementary function.
a) Let $E$ have a nonelementary right inverse.
$\ $ $E$ has a right inverse iff $E$ is surjective.
$\ $ Set $F(x)=E^{-1}(G(x))$.
$\ $ $E(F(x))=E(E^{-1}(G(x)))=G(x)$ is elementary.
b) Let $F$ have an elementary left inverse.
$\ $ $F$ has a left inverse iff $F$ has a nonempty domain and is injective.
$\ $ Set $E(x)=G(F^{-1}(x))$.
$\ $ $E(F(x))=G(F^{-1}(F(x)))=G(x)$ is elementary.
$ $
I don't know if the cases a and b are the only cases where $E(F(x))$ is elementary.
$ $
2.)
Let $E$ have an elementary left inverse.
$E$ has a left inverse iff $E$ has a nonempty domain and is injective.
$$E(F(x))=R(x)$$
$$F(x)=E^{-1}(R(x))$$
Because $E^{-1}$ is elementary and $F$ is nonelementary, and the composition of two elementary functions would be elementary again, $R$ must be nonelementary.
$ $
Bijective functions are treated in MathStackexchange: Which kinds of compositions of invertible elementary and nonelementary functions are elementary?
$\begingroup$
$\endgroup$
1
I am not quite sure what is meant by "non-elementary function", but if you have a non-elementary function $F$ such that $$ F(x)\in x+2\pi\Bbb Z\qquad\qquad\forall x\in\Bbb R $$ then $$ \cos(F(x))=\cos(x)\qquad\qquad\forall x\in\Bbb R, $$ i.e. $\cos\circ F=\cos$.
ADDED: In fact non elemetary functions $F$ as above are easy to construct. Let $G$ be just any sufficiently general non-elementary function and let $$ F(x)=x+2\pi\lfloor G(x)\rfloor,\qquad\forall x\in\Bbb R. $$
answered Feb 5, 2019 at 19:26
-
1