How can we define the relations?

[mathmari]

Hey!

Consider the ring $R=\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$.

I want to define the relation $e^x-1 \mid e^{kx}-1$ (where $k \in \mathbb{Z}$) in the language $\{+, \cdot , \frac{d}{dx} , 0, 1\}$, so we can use only these operations, the addition, the multiplication and the derivation, and all the terms come from these operations and from $0$ and $1$.

Since in the language we don't have the symbol $e^x$, I thought that this is the solution of the differential equation $y_1'=y_1$ when we apply the condition $y_1(0)=1$.

And we get $e^{kx}$ from the differential equation $y_2'=ky_2 , \ y_2(0)=1$.

So we have $$e^x-1 \mid e^{kx}-1 \Leftrightarrow \exists y_1 , y_2 \in R : y_1'=y_1 \land y_1(0)=1 \land y_2'=ky_2 \land y_2(0)=1 \land y_1-1 \mid y_2 -1$$

The relation $y_1-1 \mid y_2 -1$ can be defined in the language as follows: $$\exists h \in R : (y_1-1 )h=y_2 -1$$ right? But how can we define in the language the relations $$y_1(0)=1 \text{ and } y_2(0)=1$$ ? (Wondering)

I thought to use something like $y_1(0)=1 \Leftrightarrow \exists F: xF=y-1$, but I am not sure...

 

[Jhero]

Can you help me?Yes, you can define the relations $y_1(0) = 1$ and $y_2(0) = 1$ in the language $\{+, \cdot , \frac{d}{dx} , 0, 1\}$ as follows:$y_1(0)=1 \Leftrightarrow \exists F : xF + F = y_1-1$$y_2(0)=1 \Leftrightarrow \exists G : xG + G = y_2-1$