Is the exponential function semi-algebraic? Recall the following definitions:


*

*We say a set $E\subseteq\mathbb{R}^n$ is semi-algebraic if there exist real polynomials $g_{ij},h_{ij}:\mathbb{R}^n\rightarrow\mathbb{R}$ such that


$$E=\bigcup_{j=1}^p\bigcap_{i=1}^q\{x\in\mathbb{
R}^n:g_{ij}(x)=0\text{ and }h_{ij}(x)<0\}.$$


*

*A function $f:\mathbb{R}^n\rightarrow(-\infty,\infty]$ is called semi-algebraic, if its graph
\begin{equation*}
\{(x,y)\in\mathbb{R}^{n+1}:f(x)=y\}
\end{equation*}
is semi-algebraic. 


Literature says real polynomials are semi-algebraic, which to me is a natural result. To further understand this concept, I am wondering the following:


*

*Is the exponential function $x\mapsto e^x$ semi-algebraic? 


Unfortunately I have no idea of how to prove or disprove it, so any hint or comment will be appreciated. Thanks a billion!
Update: I am an optimizer and optimization people care about this concept because semi-algebraic functions enjoy Kurdyka-\L{}ojasiewicz property, a key assumption in many convex/non-convex optimization problems. 
 A: If $e^x$ were semi-algebraic, its graph would be a union of finitely many semialgebraic sets defined by $\{(x,y)\in\Bbb R^2\mid p_i(x,y)=0\}$. By taking the products of all the $p_i$, this means that there is a polynomial $p(x,y)$ which vanishes on all points of the form $(x,e^x)$. Expanding and collecting terms, we get $\sum_{j=1}^{n} h_j(x)e^{jx}=g(x)$ for some polynomials $g,h_j$. Taking the derivative $\deg_x(g)+1$ times, we see that we would have an equation of the form $\sum_{j=1}^n e^{jx}q_j(x)=0$ holds on $\Bbb R$ with $q_j(x)$ nonzero polynomials. But this is impossible: $e^x\neq 0$, and a nonzero polynomial only has finitely many roots.
There are also plenty of other contradictions we could derive depending on how much semialgebraic geometry you know. For instance, Lojasiewicz's inequality states that if $f,g:K\to \Bbb R$ are continuous semialgebraic functions on a compact semialgebraic set $K$ so that $f^{-1}(0)\subset g^{-1}(0)$, then we have that there exist $C,N>0$ so that $$|f(x)|\geq C|g(x)|^N$$ for all $x\in K$. Taking $K=[-1,1]$, $f=e^{-1/x^2}$ and $g=|x|$, we see that this inequality is not satisfied, but $f$ would be semialgebraic if $e^x$ was.
