MHB What could we change so that EXP(C) is a ring?

[mathmari]

Hey!

What could we change at the following definition so that $\text{EXP}(\mathbb{C})$ is a ring?
"We define EXP($\mathbb{C}$) to be the the set of expressions
\begin{equation}\label{a}
a=\alpha _0+\alpha _1e^{\mu_1z}+\dots +\alpha _Ne^{\mu_Nz}
\end{equation}
(beyond the `zero function', $0$, which we will consider to be also an element of EXP($\mathbb{C}$)),
where $\alpha_0, \alpha _1,\dots, \alpha_N\in \mathbb{C}\setminus \{ 0\}$ and $\mu_i\in \mathbb{C}\setminus \{ 0\}$; in writing such an expression we will always assume that the $\mu_i$ are pairwise distinct." Do we maybe have to assume that $\alpha_0, \alpha _1,\dots, \alpha_N\in \mathbb{C}$ and not $\alpha_0, \alpha _1,\dots, \alpha_N\in \mathbb{C}\setminus \{ 0\}$ ?

 

[Evgeny.Makarov]

Why isn't $\text{EXP}(\mathbb{C})$ a ring as it is?

 

[mathmari]

As it is now we have that $$1+e^{z} \in \text{EXP}(\mathbb{C}) \text{ and } -1+e^z \in \text{EXP}(\mathbb{C})$$ but $$(1+e^z)+(-1+e^z)=2e^z\notin \text{EXP}(\mathbb{C})$$

So, does the definition have to be as follows?

"We define EXP($\mathbb{C}$) to be the the set of expressions
\begin{equation}
a=\alpha _1e^{\mu_1z}+\dots +\alpha _Ne^{\mu_Nz}
\end{equation}
where $\alpha_i \in \mathbb{C}$ and $\mu_i\in \mathbb{C}$; in writing such an expression we will always assume that the $\mu_i$ are pairwise distinct."

 

[Evgeny.Makarov]

"We define EXP($\mathbb{C}$) to be the the set of expressions
\[
a=\alpha _0+\alpha _1e^{\mu_1z}+\dots +\alpha _Ne^{\mu_Nz}
\]
(beyond the `zero function', $0$, which we will consider to be also an element of EXP($\mathbb{C}$)),
where $\alpha_0, \alpha _1,\dots, \alpha_N\in \mathbb{C}\setminus \{ 0\}$ and $\mu_i\in \mathbb{C}\setminus \{ 0\}$

I think the definition does not require that all terms $\alpha _0$, $\alpha _1e^{\mu_1z}$, ..., $\alpha _Ne^{\mu_Nz}$ have to be present. If any of them is present, then its coefficient $\alpha_i$ has to be different from 0.

 

[mathmari]

I think the definition does not require that all terms $\alpha _0$, $\alpha _1e^{\mu_1z}$, ..., $\alpha _Ne^{\mu_Nz}$ have to be present. If any of them is present, then its coefficient $\alpha_i$ has to be different from 0.

But according to the definition of post #1, does $e^z$ belong to $\text{EXP}(\mathbb{C})$ ? We have that $\alpha_0=0$.

 

[I like Serena]

I think the definition does not require that all terms $\alpha _0$, $\alpha _1e^{\mu_1z}$, ..., $\alpha _Ne^{\mu_Nz}$ have to be present. If any of them is present, then its coefficient $\alpha_i$ has to be different from 0.

But according to the definition of post #1, does $e^z$ belong to $\text{EXP}(\mathbb{C})$ ? We have that $\alpha_0=0$.

Hey mathmari! (Wave)

I believe the definition, as it is now, explicitly states that all $\alpha_i$ up to some $\alpha_N$ have to be distinct from zero.

So what can we change? Indeed, drop that requirement and allow all $\alpha_i$ to be zero.
Moreover, $0$ is then automatically included instead of having to make an exception for it.

So the question is, will that get us that for any $a,b,c \in R$:
$$a+b\in R, (a+b)+c=a+(b+c), a+0=a, \exists (-a):a+(-a)=0, a+b=b+a$$
$$ab\in R, a(bc)=(ab)c, a\cdot 1=1\cdot a =a$$
$$a(b+c)=ab+ac, (a+b)c=ac+bc$$
? (Wondering)

 

[Evgeny.Makarov]

I agree that the original definition seems to require that $\alpha_0\ne0$. Whether this is intentional or not is hard to say. It is possible that the idea was to have elements of the form $\sum_{i=1}^n\alpha_ie^{\mu_iz}$ where $n\ge0$, $\alpha_i\ne0$ and $\mu_i\ne\mu_j$ for $i\ne j$. When $n=0$ this expression, by definition, is $0$. One of $\mu_i$ may be equal to 0; then this term is a non-zero constant. The benefit of such definition is uniqueness of representation. This uniqueness is lost if we allow all $\alpha_i$ to equal 0.

 

[I like Serena]

I agree that the original definition seems to require that $\alpha_0\ne0$. Whether this is intentional or not is hard to say. It is possible that the idea was to have elements of the form $\sum_{i=1}^n\alpha_ie^{\mu_iz}$ where $n\ge0$, $\alpha_i\ne0$ and $\mu_i\ne\mu_j$ for $i\ne j$. When $n=0$ this expression, by definition, is $0$. One of $\mu_i$ may be equal to 0; then this term is a non-zero constant. The benefit of such definition is uniqueness of representation. This uniqueness is lost if we allow all $\alpha_i$ to equal 0.

Ah. Indeed!
So it would also suffice if we only allow $\alpha_0$ to be $0$, or if we allow $\mu_i$ to be $0$ (in which case we can leave out $\alpha_0$ completely).
All other $\alpha_i$ can remain distinct from $0$.

 

[I like Serena]

The benefit of such definition is uniqueness of representation. This uniqueness is lost if we allow all $\alpha_i$ to equal 0.

Turns out it's not unique either way.
For instance $2\cdot e^{i(\pi/2)z} = -2\cdot e^{-i(\pi/2) z}$.

 

[mathmari]

Why do the elements have to be represented in an unique way? (Wondering)

 

[I like Serena]

Why do the elements have to be represented in an unique way? (Wondering)

It's not required for a ring.
But it would help if we want to deduce stuff based on a specific $\mu_i$.
Then the powers should be independent.

 

[I like Serena]

Turns out it's not unique either way.
For instance $2\cdot e^{i(\pi/2)z} = -2\cdot e^{-i(\pi/2) z}$.

Nevermind. I made a mistake with the $z$. They are not equal after all.

Why do the elements have to be represented in an unique way? (Wondering)

We need it to be able to say that if for any $z \in \mathbb C$ and $\alpha_i,\beta_i \in \mathbb C$ we have:
$$\sum \alpha_i e^{\mu_i z} = \sum \beta_j e^{k_j z}$$
that then:
$$\begin{cases}
\alpha_i = \beta_j &\text{if } \mu_i = k_j\\
\alpha_i = 0 &\text{if } \forall j: \mu_i \ne k_j\\
\beta_j = 0 &\text{if } \forall i: \mu_i \ne k_j\\
\end{cases}$$

And anyway, it's just neat to have a unique representation for every element in the ring.