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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?

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    $\begingroup$ Can't you define it as the element $y$ such that $dy/dx = y$? $\endgroup$ Dec 3, 2015 at 17:01
  • $\begingroup$ But do we not get from $\frac{dy}{dx}=y$ the elements $Ce^x$? Do we have to get then rid of $C$ ? @YoavKallus $\endgroup$
    – Mary Star
    Dec 3, 2015 at 17:31
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    $\begingroup$ Yes, you do. I was assuming that you need to define $e^x$ up to scalars. Note that your language is invariant under $x\mapsto x+t$, so, unless I missed something, $e^x$ cannot be defined. $\endgroup$ Dec 3, 2015 at 17:44
  • $\begingroup$ What do you mean by "defining $e^x$ up to scalars" ? @YoavKallus $\endgroup$
    – Mary Star
    Dec 3, 2015 at 17:49
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    $\begingroup$ Precisely what you wrote above, $\{Ce^x:C\in\mathbb{C}\}$. $\endgroup$ Dec 3, 2015 at 17:53

1 Answer 1

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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.

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  • $\begingroup$ So it isn't possible to define the relation $$e^x-1 \mid e^{kx}-1$$ in this language, right? To do that do we have to add the symbol $e^x$ in the language? $\endgroup$
    – Mary Star
    Dec 3, 2015 at 19:10
  • $\begingroup$ I don't know what you mean by the relation you wrote down. $\endgroup$ Dec 3, 2015 at 20:29
  • $\begingroup$ For a reduction I need the following implication: $$k\in \mathbb{Z} \Leftrightarrow e^x-1\mid e^{kx}-1$$ Can we define this in the language ? $$$$ P.S. I want to reduce the theory of the structure $$\left (\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}] ; +, \cdot , \frac{d}{dx} , 0, 1\right )$$ into the theory of the structure $$\left (\mathbb{Z}; +, \cdot , 0, 1\right )$$ $\endgroup$
    – Mary Star
    Dec 3, 2015 at 22:17
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    $\begingroup$ @MaryStar I believe you can define the set of pairs $(f,g)$ such that $f = c e^x$ and $g = c^k e^{kx}$ for some (common) $c$. This will allow you to define a relation equivalent to the one you want. I.e. suppose the definition is the formula $\phi_k(f,g)$, then you want the formula $\psi_k$ written as $\forall f\forall g (\phi_k(f,g)\to(\exists h((f-1)\cdot h=(g-1)))$ $\endgroup$ Dec 4, 2015 at 0:01
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    $\begingroup$ Yeah, but without the quantifiers. $\endgroup$ Dec 4, 2015 at 0:44

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