We consider the ring $\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$ and the language $L=\{+, \cdot , \frac{d}{dx} , 0, 1\}$. 


The ring consists of elements of the form $$\sum\_{i=0}^N \alpha\_i e^{\lambda\_i x}$$ where $\alpha\_i , \lambda\_i \in \mathbb{C}$. 


In the language there is no symbol $e^x$. 


When we want to write a formula in the structure $$\left (\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}] ; +, \cdot , \frac{d}{dx} , 0, 1\right )$$ can we use the symbol $e^x$ because it is an element of the ring? 


Or do we have to define it somehow using the operations of the language? 



Can we use this symbol?
Note that the transformation $x\mapsto x+t$ leaves all relations defined by your language invariant.


Namely,


1. $\sum{a\_i e^{q\_i x}}+\sum{b\_i e^{r\_i x}}=\sum{c\_i e^{s\_i x}}$ iff $\sum{a\_i e^{q\_i (x+t)}}+\sum{b\_i e^{r\_i (x+t)}}=\sum{c\_i e^{s\_i (x+t)}}$
2. $(\sum{a\_i e^{q\_i x}})\cdot(\sum{b\_i e^{r\_i x}})=\sum{c\_i e^{s\_i x}}$ iff $(\sum{a\_i e^{q\_i (x+t)}})\cdot(\sum{b\_i e^{r\_i (x+t)}})=\sum{c\_i e^{s\_i (x+t)}}$
3. $(d/dx)\sum{a\_i e^{q\_i x}}=\sum{b\_i e^{r\_i x}}$ iff $(d/dx)\sum{a\_i e^{q\_i (x+t)}}=\sum{b\_i e^{r\_i (x+t)}}$
4. $\sum{a\_i e^{q\_i x}} = 1$ iff $\sum{a\_i e^{q\_i (x+t)}}=1$
5. $\sum{a\_i e^{q\_i x}} = 0$ iff $\sum{a\_i e^{q\_i (x+t)}}=0$


Therefore, any formula in this language defines a set that is invariant under this transformation. The set $\{e^x\}$ is not and therefore not definable. In contrast, the set $\{a e^x: a\in \mathbb{C}\}$ is.


